Is using the same IV in AES similar to not using an IV in the first place? So if I understand how an IV works with AES, I'm supposed to generate a different IV for every message because using only a key, I will get the same encryption if the message was encrypted twice (which is not secure) thus we use the IV which is some kind of a salt (some random bytes added to the encrypted message so 2 messages with the same value won't have the same encryption).

Now my question, is using the same IV for all of my messages the same as not using an IV at all? Or is it slightly better?
Lets see if I can clarify things for you.

For one, the IV is not specifically related to AES at all.  AES is a keyed invertible transform from a 128 bit value to a 128 bit value; that's all it can do.  Now, if you just happen to have a 128 bit value that you want 'encrypted' into a 128 bit ciphertext, well, you can just use AES as is.

However, we typically want to other things than that; we may have plaintexts which aren't exactly 128 bits long; we might want nondeterministic encryption (so that if we encrypt the same value twice, it's not obvious that we did so); we might want integrity checking on our ciphertext as well.

To allow us to do those things, we use a Mode of Operation; that's a construction that uses AES (or some other block cipher) to perform these more generally useful operations.  It is these modes of operation which may (not all do) take an IV.

At the moment, the most popular modes of operation are CBC and GCM; they also have very different requirements to their IVs.

So, in that context, I'll answer your questions:


  I'm supposed to generate a different IV for every msg because using only a key I will get the same encryption if the message was encrypted twice


That's not the complete story; yes, if you use the same IV, same message and same key (and same AAD for GCM), you'll get the same ciphertext, but that's not the full reason.

For CBC, if you use the same IV, there are two potential problems; one is that you'll leak how the two plaintexts are related.  For example, if the two plaintexts have the same first 16 bytes, and differ in the second 16 bytes, that'll be obvious to the attacker.  The other is that if the attacker can predict the IV (which he can if you use the same one repeatedly), and if he can introduce his own plaintext (which can happen in some scenarios), then he can use the encryption operation as an oracle (and thus deduce the value of low entropy plaintext blocks encrypted by the same key).

For GCM, if you use the same IV, there are two different potential problems (both of which are far more serious); for one, he can deduce the bit-wise xor of the two plaintexts (which is often enough for him to deduce those plaintexts).  Worse, it allows the attacker to deduce the internal authentication value, and thus allow him to modify ciphertexts (and thus the resulting plaintexts) without being detected.


  Now my question, is using the same IV for all of my messages is the same of not using IV at all? Or is it slightly better? 


Well, for neither mode, you don't have an option of 'not using an IV'; when you perform a CBC or GCM operation, an IV must be there (for GCM, you can have a zero length IV; that's not quite the same as 'not having an IV').

And, you didn't ask, but I'll state the general wisdom anyways:


For CBC mode, the IV really ought to be unpredictable to the adversary.  Now, you can pick it randomly each time, or you can use (say) an AES encryption operation with the same key to generate it (which the attacker can't predict, as he doesn't have the key); both works.
For GCM mode, it doesn't matter in the least whether the IV is predictable to the adversary.  What's critical is that you use a different one each time.  If you can manage a counter (e.g. if you restart the program, you'll pick a fresh key), then using IV=0 for your first message, IV=1 for your second, etc, is the obvious approach.