No, this is not possible. If you change a single bit in a block that is put through the cipher then all bits change with a certainty of 50% (each) due to the avalanche effect. So even though only one bit will be changed in the next block, the current block will still "completely" change.
Differently said, AES is a block cipher and a keyed block cipher is a pseudo random permutation. If you change even a single bit you get another mapping to an entirely different value for that block. Only one block leads to the same plaintext block, there is no redundancy. Because of the pseudo-randomness you get an entirely different plaintext block, no matter how many bits you change.
Some padding modes are redundant though (the most common one, PKCS#7 compatible padding isn't). So there is a chance that an attacker can change the last few bytes of the plaintext. For instance, you'd have a ~ 1 / 65536 chance that you'd create an ISO 10126 padded message that is identical after unpadding (the first byte identical and the last byte indicating 15 padding bytes).
Obviously using an authentication tag / message authentication code (MAC) over the ciphertext should be considered to be secure against any of these attacks.