as mentioned in the title, and read some past discusstion on AES collison here.
it shows it's possible that, AES(k1, m1)=AES(k2, m1)
just wondering what condition those 2 keys need to meet. Thanks.
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AES is (as far as is known) a very good approximation of a Pseudo-Random Permutation. There's no discernible pattern to how keys transform plaintext into ciphertext blocks. If there were, that would be usable in a Known-Plaintext attack to recover the key given some number of plaintext-ciphertext pairs. That would be a massive security break for AES.
AES used in a Mode of Operation, however, can be different. There extra inputs aside from the key and plaintext block are allowed. This could make it easier to find a collision, or it might make it harder (eg Merkle-Damgaard mode turns a block cipher into a collision-resistant hash function).
With only 6 colliding bytes needed it's likely fast enough to simply brute-force two keys.