Let $R$ be a commutative Noetherian ring. Given an $R$-module $M$ and a ring homomorphism $\phi:R\to R$, let $\phi^* M$ be the $R$-module whose underlying abelian group is the same as $M$ but the $R$-linear structure is given by $r\cdot m=\phi(r)m, \forall r\in R, m \in \phi^* M$.
Now let $(R,\mathfrak m)$ be, moreover, local and $\phi:R\to R$ be a ring homomorphism such that $\phi^n(\mathfrak m)\subseteq \mathfrak m^n$ and $(\phi^n)^*R$ is a finitely generated $R$-module for all $n\geq 1$ (here, $\phi^n$ denotes $n$-fold composition). If $(\phi^e)^* R$ is a free $R$-module for some $e>0$, then is $(\phi^s)^* R$ a free $R$-module for infinitely many $s>0$ ?
The functor $f^*$ has both adjoints, so it preserves all limits and colimits, so in particular if $(\phi^s)^*(R)$ is a free module with generators indexed by $I$, then $(\phi^{ks})^*(R)$ is free with generators indexed by $I^k$.