If a permutation $f$ is not one way, what can we say about $f^{p(n)}$?
If a permutation $f$ is not one way, we can not conclude about the one-wayness of $f^{p(n)}$. In fact, even $f^2$ could be one-way, if there are one-way length-preserving permutations that is.

Constructive proof: assume $g:\{0,1\}^*\rightarrow \{0,1\}^*$ is a length-preserving OWP, not invertible starting with rank $m$. Define $f$ from $g$ such that, for any bitstring $x$,

$$\begin{align}f(x\mathbin\|0)&amp;=x\mathbin\|1\\f(x\mathbin\|1)&amp;=g(x)\mathbin\|0\end{align}$$

$f$ is a permutation. It is trivially invertible for half its inputs of any fixed width, thus is not a OWP.

$f^2$ is also a permutation, with

$$\begin{align}f^2(x\mathbin\|0)&amp;=g(x)\mathbin\|0\\f^2(x\mathbin\|1)&amp;=g(x)\mathbin\|1\end{align}$$

An hypothetical algorithm that inverts $f^2$ starting at width $n&gt;m$ could be used to invert $g$ at width $n-1\ge m$. With $g$ being a OWF, there is no such algorithm. Hence $f^2$ is one-way.