Is a random permutation "hiding"? Let's say I tell you $G: \mathbb{F} \rightarrow \mathbb{F}$ is a random permutation (some finite field). Does that mean:


$G$ is one-way, so that if I give you $G(x)$, it is infeasible to determine the pre-image $x$?
Is $G$ hiding in the sense that you can't figure out anything about $x$ given $G(x)$? For example, you could determine if $x$ were even/odd?


If the second question is true, how do I capture this in a game?
If $G$ is only known to be a random permutation, and one $G(x)$ is given, that gives no information whatsoever about $x$ (it can be any element of $\mathbb F$ with equal likelihood if $G$ is uniformly random). So I would say $G$ is one-way and hiding in the sense of the question.

Things change when for the same $G$ it is considered several given $G(x)$ and perhaps $x$. In particular, knowing $G(x)$ and $G(x')$ we can tell if $x=x'$, since that is equivalent to the testable $G(x)=G(x')$.

Capturing the first situation in a game is easy: change the permutation whenever its input or output changes. But that's bound to give dull results: we're not using most of what the permutation is good for.

To capture the second situation, define the permutation early in the game, and make several rounds with $x$ or $G(x)$ varying. It's possible to establish things like: for appropriately large $\mathbb F$, given a $G(x)$ that was not previously disclosed, the computational advantage in predicting $x$ is negligible.