This question is related to the one I have asked yesterday.
For an integer $m\geq 1$, consider the $m$th cyclotomic polynomial $\Phi_m(X)\in\mathbb Z[X]$. For an integer $r\geq 1$, we can find polynomials $F_1,\dots,F_n\in\mathbb Z_{p^r}[X]$ $\big($with $\mathbb Z_{p^r}:=\frac{\mathbb Z}{(p^r)}\big)$ for which $$\Phi_m\equiv F_1\cdots F_n\bmod p^r$$ and for which the $F_i\bmod p$ are the irreducible factors of $\Phi_m\bmod p$ (they are distinct and all have the same degree). The $F_i$ can be constructed using Hensel lifting, which was basically the content of the linked question.
We have an isomorphism of rings $$\frac{\mathbb Z_p[X]}{(F_1\mod p)}\ni f(X)\mapsto f(X^p)\in\frac{\mathbb Z_p[X]}{(F_1\mod p)},$$
which follows from the fact that for $f(X)\in\mathbb Z_p[X]$, we have $f(X^p)=f(X)^p$ (because of Fermat's Little Theorem: $a^p\equiv a\bmod p$ for every $a\in\mathbb Z$).
Apparently, we also have an isomorphism of rings given by $$\frac{\mathbb Z_{p^r}[X]}{(F_1)}\ni f(X)\mapsto f(X^p)\in\frac{\mathbb Z_{p^r}[X]}{(F_1)}.$$
In other words, for $f(X)\in\mathbb Z_{p^r}[X]$, we have the equivalence $$F_1\text{ divides } f(X)\iff F_1\text{ divides } f(X^p).$$
In this case, we cannot use Fermat's Little Theorem because $a^p\not\equiv a\bmod p^r$ does generally not hold. Further, we cannot really argue with the notion of irreducible polynomials (this notion requires us to work in an integral domain). What other property can we use to argue that this is an isomorphism?
It is not true if $p|m$, try with $p^r=9,m=3,F_1=\Phi_3=X^2+X+1$, it divides $X^3-1$ so $\zeta^3=1$ in $\Bbb{Z}/(9)[\zeta]/(F_1(\zeta))$. And $1$ is not a root of $F_1(Y)\in \Bbb{Z}/(9)[\zeta]/(F_1(\zeta))[Y]$. ie. no ring endomorphism sends $\zeta$ to $\zeta^3$.
Now assume that $p\nmid m$.
Because $F_1$ has integer coefficients, $\zeta^p$ is a root of $F_1(Y)\in \Bbb{Z}/(p)[\zeta]/(F_1(\zeta))[Y]$
By Hensel lemma it lifts to some root $\alpha$ of $F_1(Y)\in \Bbb{Z}/(p^r)[\zeta]/(F_1(\zeta))[Y]$
Now $$\Phi_m(Y)=\prod_{k=1,\gcd(k,m)=1}^m (Y-\zeta^k) \in \Bbb{Z}/(p^r)[\zeta]/(F_1(\zeta))[Y]$$ Because $\Phi_m$ is separable $\bmod p$ you get that $\Phi_m(\alpha)=0 \implies \alpha = \zeta^k$ for some $k$, obviously it must be $p$.
So we proved that $\zeta^p$ is a root of $F_1$ and hence $\zeta\mapsto \zeta^p$ is an automorphism of $\Bbb{Z}/(p^r)[\zeta]/(F_1(\zeta))$.
Honnestly this is a bit messy without $p$-adic numbers (and their finite extensions), they were invented exactly for this kind of question.