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:
C1 = M1 ⊕ K
C2 = M2 ⊕ 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:
C1 ⊕ C2 = M1 ⊕ K ⊕ M2 ⊕ K
which, after cancelling out the two K values (since x ⊕ x ≡ 0) gives us:
M1 ⊕ M2
This allows an attacker to know which bits of M1 are equal to M2. 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 SEa great answer on Crypto SE about exactly how this attack works.
You should always use a unique IV per message to avoid this problem.
