The RSA-Assumption says that $(GenSP,F,SampleX)$ is one-way. So if we initialize an instance of RSA $(n,e), (n,d)$ and fairly forget the secret-key and SampleX uniformly distributed over $X, F = x^e \bmod N$ should be one-way.
Now it is also known that Injective functions imply collision resistance, but not one-way of course.
So far, we have pre-image resistance and collision resistance. And the second pre-image resistance should be given if we take $x$ only from $\{0,\ldots,n-1\}$.
We talk about deterministic RSA, so no randomized process is involved with random-padding. Therefore our F is also deterministic.
Am I missing something?
(e, N)
pair (which would have to be standardized, if we were to use RSA as a hash function), that you truly did throw away the corresponding private key, respectively factorization of the modulus. $\endgroup$