Let $\Phi : F[x] \to F[x]$ be an isomorphism such that $\Phi(a) = a$ for every $a \in F$. Show that $f(x)$ is irreducible in $F[x]$ if and only if $\Phi(f(x))$ is irreducible in $F[x]$.
Any help on how to go about starting my contrapositive proof please
$\Phi$ is an isomorphism. If $f = g h$, then $\Phi(f) = \Phi(g)\Phi(h)$ and vice versa. Since $\Phi$ maps the degree zero polynomials (the constants) to themselves, $g$ and $\Phi(g)$ are either both constants or both non-constants. Likewise for $h$ and $\Phi(h)$. Apply definition of irreducible. Done.