**You must not do this.** AES in CTR mode turns it into a stream cipher, such that AES is turned into a cryptographic pseudorandom number generator (PRNG) which generates a sequence of pseudorandom bits to be used as a keystream. This output keystream is simply xor'ed with the plaintext stream to produce ciphertext. Using the same key and IV produces the same keystream each time, so by re-using the key and IV you're essentially using a many-time pad. So, you get: > C<sub>1</sub> = M<sub>1</sub> ⊕ K > > C<sub>2</sub> = M<sub>2</sub> ⊕ K Where *C* is ciphertext, *M* is the plaintext message, *K* is the keystream produced by AES-CTR, and ⊕ denotes an exclusive-or (xor) operation. By computing the xor of those two ciphertexts, you get: > C<sub>1</sub> ⊕ C<sub>2</sub> = M<sub>1</sub> ⊕ K ⊕ M<sub>2</sub> ⊕ K which, after cancelling out the two *K* values (since x ⊕ x ≡ 0) gives us: > M<sub>1</sub> ⊕ M<sub>2</sub> This allows an attacker to know which bits of M<sub>1</sub> are equal to M<sub>2</sub>. If you know any bits of one message (known plaintext), you can then recover the corresponding bits from the other message. You can also employ techniques such as crib-dragging to fully decrypt the entire message. There's [a great answer on Crypto SE](https://crypto.stackexchange.com/questions/2249/how-does-one-attack-a-two-time-pad-i-e-one-time-pad-with-key-reuse) about exactly how this attack works. You should always use a unique IV per message to avoid this problem.