I want to prove that $x\mapsto x^i$, $i\in [r-1]$ gives automorphism over the ring $\mathbb{F}_p[x]/(\Phi_r(x))$, $\Phi_r(x)$ being the $r$th cyclotomic polynomial, $r$ being a prime.
I was able to derive that this gives an endomorphism, as $\Phi_r(x)\mid \Phi_r(x^i)$, so root moves to root and $1,x$ being a generator, of the $\mathbb{F}_p$ algebra, we can derive the endomorphism. But why it will be an automorphism that is my question.
Note- $\Phi_r(x)$ is assumed reducible in $\mathbb{F}_p$ as if it is irreducible then the question is trivial.
It is an endomorphism with an inverse: $x^r=1$ so $(x^i)^j=x$ where $ij=1\bmod r$