Well, the only thing that mirrors into anything on that grid is the 2 mirroring into a 5. I guess it is flipped vertically since the 3 appears normally. (It is also possible that the numbers that go into some spots need to be mirrored and/or swapped into the other spots as they flip into the next grid, but for now try to solve the first grid.)

It is still the case that 4 and 2 must go in the first two cells of the bottom row of the region containing 1 and 3.

In the upper left region, the 2 can go in either the top left or middle left. Suppose it goes into the middle left. The cell below the 2 cannot contain 2, 5, or 4 because of their appearance elsewhere, nor can it contain 1 or 3 because they are adjacent to 2, so it must contain 6. Now the two remaining cells in the top row must contain 4 and 5 in some order, but these cannot be adjacent to each other so the assumption must be wrong.

Instead, 2 goes in the upper middle cell of the upper left region. 1 and 3 are forced into the lower corners of this region, with 3 on the left. 4 goes in the upper right to avoid being next to 3, and 5 goes in the lower middle to avoid being in line with another 5. This also forces the 5 and 6 in the upper right cell (5 in top row). So the first two rows of the grid look like:

624153
351426

Now in the middle left region, 2 is blocked from the middle column by the other 2, and also from the upper corners because that would place it next to 1 or 3. So it goes in a bottom corner and the 2 in the middle right region must go in the top row.

There are two cases now. If 2 goes in the upper left, then 3 must go in the bottom middle, 4 must go in the upper right, 1 must go in the bottom right, 5 in the bottom left, and 6 in the upper middle.

264
531

If 2 goes in the upper left, 1 must go in the bottom middle, and 3 must go in the bottom left. Now 5 must go in the bottom right because placing it in the top row would force it to go next to a 4 or 6 in the other empty top row cell. So 6 goes in the upper left and 4 in the upper middle.

642
315

Each of these forces a corresponding solution in the lower right region. 2 and 5 must go in the column which does not have a 5 yet, and 5 must go on top since the bottom row already has a 5. In the first case a 1 and 4 must go in the middle column and the 1 must go on top by adjacency. Then the 3 and 6 also have a unique solution by adjacency. The other case also has a unique solution forced by adjacency.

264
531
315
642

642
315
531
264

In any case, 3 and 1 must go in the remaining bottom row squares, with 1 in column 2 because of the existing 1 in column 3. The numbers 2, 4, 6 are forced into the top row of this region by like numbers and adjacency with the 5,1,3 below them. This eliminates the second case above as a valid solution since it places 6 next to 5. All the numbers in the last region are forced.

624153
351426
135264
462531
246315
513642

Lucky for us, the two numbers in the gray cells are the same, so there is no worry they need to be swapped. Copy them into the next grid. We can also copy a 3 backward to the previous grid.

In the grid following the solved one, there are chains of consecutive numbers involving each 4. So a 5 and 3 appear adjacent to each one, and the other adjacent number must be 2 or 6 because it is not consecutive with the other 3 or 5 (which must thus be the other of 3 and 5). So each of these regions contains a chain of 4 consecutive numbers, either 2,3,4,5 or 6,5,4,3.

This grid is symmetric, so it will probably have a symmetric solution unless an asymmetry is introduced by the numbers in the center shaded cells in the next grid. Suppose that the two sides of this grid fill in differently, with a 3 in the same region as one 4 and a 5 in the other. This forces the other of 5 and 3 to lie on the same row in each region as the 4, so there is 4,5,3 in one region and 4,3,5 in the other (the one on the right reading backward). So 1, 2, 6 must go in the other row with no adjacencies. This means 6 must go in the middle to prevent 1 and 2 from being adjacent, but on one side the 6 is going to be adjacent to a 5 where it shouldn't be. So the two consecutive sequences through the 4s are the same.

Suppose the numbers next to the 4s are 3s. Now 2 must go in a top corner of the left region and a bottom corner of the right. The 1 must go in the other of these corners, because any other placement would put it next to the 2 or would force the 5 and 6 to be adjacent, while none of the remaining cells in these regions are allowed to be consecutive. Since there is a 6 in the middle column in this case, the 5 must go in the middle column and the 6 goes in the inner spot of the row with 4 and 3:

xxxxxx
xxxx65
x5x634
436x5x
56xxxx
xxxxxx

Otherwise, 5 goes next to the 4.

This forces the 6 into one of the top corners of the left region. If it is directly above the 4, there is no way to place the 2 where it is not adjacent to the 1 or 3. So the 6 goes in the inner corner, The 3 goes in the middle column (else it shares a column with another 3 or forces 1 and 2 to be adjacent). Then 1 goes next to the 3 and 2 is forced.

xxxxxx
xxxx23
136254
452631
32xxxx
xxxxxx

At the top of the second column there are two consecutive digits. The case has only 1, 4, and 6 unused in this column, so it is impossible. In the former case, these are 1 and 2 in some order; 4 must go at the bottom. This 4 is next to a consecutive number which is 3 since 5 already appears in the region.

xxx34x
xxxx65
x5x634
436x5x
56xxxx
x43xxx

Now the 1 and 2 in this region are forced (1 not next to 3) and this forces all the other 1s and 2s, and then all the remaining numbers are forced. All the adjacencies come out correct, and 2s get copied into the next grid.

615342
324165
152634
436251
561423
243516

Unfortunately 2s do not work well in this spot in the next grid (which is again symmetric), as we have this same configuration where three out of 4 edges in a square are consecutive, which can (once again) only be filled by a chain of 4 consecutive numbers. Since the 2 comes at the start of the chain, it can only be 2,3,4,5 which puts the 4 in the region that already contains a 4.

Well maybe the 2s get mirrored into 5s when you copy them. Then the sequence is 5,4,3,2 which puts a 3 in the region with a 4.

xxxx4x
xx43xx
xx52xx
xx25xx
xx34xx
x4xxxx

Just to one side of the adjacent group is a smaller adjacent group (only 3 consecutive adjacent numbers, which could be all different or first and last the same). The one immediately next to the 5 can only be 1 or 3, and 1 forces a second 2 in the 3rd & 4th rows, so it is a 3, with a 4 on the outside and a 2 adjacent on the other side.

xxxx4x
x243xx
4352xx
xx2543
xx342x
x4xxxx

All the remaining entries are not consecutive with their adjacent numbers, so the 6 must go next to the 2 in the 3rd & 4th rows. The 1 goes between 6 and 4 in columns 2 and 5. Then the 5 in rows 2 and 5 is forced to not be in the same region with another 5, and everything else is forced.

351642
624315
435261
162543
513426
346153

A 1 and a 2 are copied to the next grid -- or is it a 5? And do the numbers get swapped around?

xxxxxx
xxxxxx
xxxxxx
xxxxxx
x1xx2x (or 1 and 5? or 5 and 1? or 2 and 1?)
xx3xxx

In the lower left cell, if the number coming in from the previous grid is a 1, and the 2 cannot go next to it so it must go in the lower left corner (regardless of the number given in the other region). The 4 cannot go next to the 3 so it also goes on the left, and the 5 and 6 can go either way.

41xx?x
2x3xxx

Now the given in the lower right cannot be a 2, because it would force the 3 to go into the bottom row which already has a 3. So it is a 5, and the 4 and 6 must go in the lower corners of this region. The 4 cannot go next to the 3 so the 6 does. This also forces the 5 and 6 in the left region. The 1 must go in the bottom row, and the 3 must not go next to the 4, so everything is forced.

416352
253614

Suppose the numbers copied from the previous grid get swapped. So the 1 goes in the lower right cell. Then the 2 is forced into a bottom corner, not next to the 3 so it is on the far right. Then the 3 cannot go on the bottom or adjacent to the 2 so it goes left of the 1. The 5 must go in the top row to avoid two adjacent numbers on the bottom, and once again the 4 must not go next to the 3. And in this case the given in the other cell is forced to be a 2.

x2x315
xx3642

On the left, the 1 does not go next to the 2 or in the top row, so it goes in the lower left corner. The 4 does not go next to the 3 so it goes in the upper left. And the 5 and 6 are forcesd

426315
153642

Only one of these numbers, the one on the far right, is consecutive with the number above it, and that number also has a consecutive number to its left. In the latter case, the number above the 5 must be a 6, because putting a 4 there does not allow there to be two other consecutive numbers in this column. This forces the other adjacent number to the 6 to be a 5. The 5th column also has a pair of adjacent numbers above the ones known so far, and these must be 2 and 3. So we can fill in the top numbers.

xxxx61
xxxxxx
xxxxxx
xxxx56
426315
153642

But neither pair of the unknown numbers in the 5th and 6th columns is adjacent to a consecutive number horizontally, but the 3 in column 6 can only be adjacent to either a 2 or another 3, in either case impossible. So go with the former case.

The consecutive number above the 2 can be either a 3 or a 1, but the number to its left is always a 2 because a 4 would be consecutive to the 5.

xxxx2?
416352
253614

In any case, 3 and 4 are the consecutive numbers in column 5, and 5 and 6 are the consecutive numbers in column 6, forcing a 6 and the other of 3 and 1 into the top row.

xxxx6?
xxxxxx
xxxxxx
xxxx2?
416352
253614

In the middle right cell, there are no more adjacent consecutive numbers, so the unplaced one of 1 and 3 goes in the upper left cell. That column contains a 3 already, so this is a 1 and the 3 is beside the 2. And of the consecutive 3 and 4, the 3 must not go next to the 2 (or in the region which already has a 3). Likewise the consecutive 5 and 6 go with the 5 not adjacent to the 4. The last number in the middle right region is a 5.

xxxx61
xxxx35
xxx146
xxx523
416352
253614

The two adjacent numbers in row 3 must be 2 and 3, and the 2 in the bottom row forces the 3 to the far left. In the row below, the 4 is forced into the second column, and the 6 into the first column.

xxxx61
xxxx35
325146
641523
416352
253614

The 2 and 4 go in column 4 somehow. Another 4 and 2 go into column 3, and the 4 cannot go adjacent to the 5 in column 3. The consecutive number at the top of column 2 is a 3, and the remaining numbers are forced.

534261
162435
325146
641523
416352
253614

So it is solved, twice turning a 2 into a 5 in the copying in addition to the one in the grid.

Joe