AES key equal to IV (CBC mode) Application uses AES in CBC mode with 128 bit key. In this implementation key is always same and is equal to IV.
Can attacker get the key or decrypt data? If he knows:


First and second blocks of plaintexts;
All blocks of ciphertexts.


The software mentioned above is malicious program, it encrypts all office files on user's computer. And I now attempt to help user get his data back. Reverse engineering of this program shows that key is unknown, but is always same and is equal to IV.
In this answer I'm assuming that a key is used to encrypt more than one message.


The first weakness is that CBC with fixed IV leaks if messages share a common prefix.
The second weakness is that it makes a padding oracle attack much more severe. Consider a device that knows the key and decrypts a ciphertext you send to it. While it won't tell you the plaintext, it'll leak if the padding was valid (e.g. via an error message or timing). Such a device acts as padding oracle.

For normal use of CBC with PKCS#7 padding and a known (variable or fixed) IV, such a padding oracle will allow an attacker to recover the message associated with plaintext, but it will not leak the key.

When using a fixed secret IV, it doesn't recover the plaintext of the first message block directly, rather it recovers $m_1 \oplus \mathrm{IV}$. When the attacker has access to several ciphertexts, this behaves like a many-time-pad thanks to the reused IV. Due to the redundancies of natural languages this should allow recovery of the IV given only a few ciphertexts.

If the key is reused as IV this IV recovery attack becomes a key recovery attack.


I don't expect either of these weaknesses to help a ransomware victim, since the first is only a minor leak and the second requires a padding oracle, which you probably won't have.