yes, there is a solutionfound one by hand in about 20 minutes of shuffling knights around with my finger on lichess's board setup editor on my phone. it's a fun little exercise
step 1: look for patterns with 1 digiti started by looking for patterns with 1 digit where all the digits are a knight's move apart and there is only 1 knight in each row and column. i did this manually on lichess's board setup editor
i found 5 of them. there may well be more, but i stopped here. note that all these patterns have various rotations and reflections, which doesn't change the pattern
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pattern A:
1.......
..1.....
.1......
...1....
.....1..
.......1
....1...
......1.
pattern B:
1.......
..1.....
....1...
......1.
.......1
.....1..
...1....
.1......
pattern C:
.1......
1.......
..1.....
....1...
...1....
.....1..
.......1
......1.
pattern D:
.1......
1.......
..1.....
......1.
...1....
.....1..
.......1
....1...
pattern E:
..1.....
.1......
...1....
1.......
....1...
.......1
.....1..
......1.
step 2: combine 2 of them in the grid trying to avoid symmetric digitswhen trying to find 1-digit patterns, it is noticeable how restrictive the given conditions are, so it seems likely that to find a solution for a full grid it's worth exploiting some of the symmetries of the grid, instead of trying to fit 8 of these patterns in one at a time.
so what this means is it's worth trying to fit 2 of these patterns into the grid so that no pair of horizontally or vertically symmetric cells contain a pair of digits. (example, using chess notation: b1 and g1 are row-symmetric; a4 and a5 are column-symmetric) this is because we can then fit 4 copies of this pattern using reflections into a full solution grid without any clashes.
trying to find 2 of these to fit together was done pretty much by guessing until i found one that worked, but there are some combinations that can be instantly ruled out, such as any two patterns that contain a corner cell can't be used here because then you'd have 2 corner cells, and corners are invariant under rotation/reflection so you'd end up having to fit 8 corner cells into the grid after you fit 4 copies of this pattern in, but there are only 4 corners. the same logic goes for the centre cells, and the diagonal cells (b2,b7,g2,g7) and (c3,c6,f3,f6)
after a while i found that diagonally-reflected pattern B combines with pattern E to get the combination i want:
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pattern E in 1s + pattern B diagonally-reflected in 2s:
2.1.....
.1.....2
.2.1....
1.....2.
..2.1...
.....2.1
...2.1..
....2.1.
this works because no row or column contains a 1 and 2 that are symmetric with respect to the centre line of the grid, and the diagonal cells are only represented once each for each equivalence class
step 3: apply reflectionsthis is the easy part. apply all combinations of horizontal and vertical reflections to the above partially completed grid and you get a solution
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25187364
41578632
52813746
14756823
68231475
37465281
86324157
73642518
there may be other solutions but i stop here