An RSA key is actually a key pair: a public key and a private key. The public key is usually uploaded to a server. If I understand correctly, in this situation, attacking RSA amounts to solving a math factoring problem, for which there are (currently theoretical) efficient quantum algorithms.
My question is, what if the RSA public key is not actually made public at all? Is it as easy to attack such "purely private" RSA encryption than it is to attack "normal" RSA encryption, in which the public key is known? Is the public key somehow embedded in the RSA encrypted message? If not, do the quantum algorithms need the RSA public key to work?
If purely private RSA is harder to attack, how does purely private 2048/4096 RSA encryption compare to 128/256 AES?
Edit: I understand that:
RSA is usually only used to encrypt a random "session key", which is then used for symmetric (say, AES) cryptography. E.g., in OpenPGP: https://tools.ietf.org/html/rfc4880#section-2.1 .
The extra RSA layer introduces an extra vulnerability: one can attack it by breaking either RSA or the session AES.
If the extra RSA layer uses a known public key, it is "much easier" to break that, at least in theory, using future quantum algorithms, than it is to break AES.
What I'm trying to understand is, how does "purely private" RSA + session AES compare with AES alone (assume the same AES key size).
Edit 2: To avoid unnecessary tangents, I added an extensive clarification of what I'm trying to understand here. The question there is "Is there a hacky way to remove the unnecessary RSA?". The OP here asks "Just how bad is the unnecessary RSA?"