Consider a permutation $f:\{0,1\}^*\rightarrow \{0,1\}^*$, which is not a one-way function, i.e. there exists an efficient probabilistic adversary $\mathcal{A}$ and some polynomial $q(n)$ such that for infinitely many $n$
\begin{equation} \mathrm{Pr}[\mathcal{A}(f(x)) = x] > \frac{1}{q(n)}, \end{equation} where the probability is over the internal randomness of $\mathcal{A}$ and $x$ drawn uniformly at random from $\{0,1\}^n$.
I would now like to prove that $f^{p(n)}$ is not a one way function, for any polynomial $p(n)$. Is this true for all permutations $f$ which are not one way?
My intuition was that this would be true, and that one could prove this by showing that $\tilde{\mathcal{A}} = \mathcal{A}^{p(n)}$ is a suitable adversary for $f^{p(n)}$. However, I can't get this to work. Is there a clever way to prove this, or is the statement I am trying to prove actually false?