Let $F$ be a field and $f: F[x] \to F[x]$ with $f\left(p\right) := p\left(ax\right)$. In fact, $f$ is a isomorphism and this is a elementary proof, with the "one-to-one" part of it just by showing that $f(g) = f(h) \Rightarrow g=h$ and the properties of a homomorphism did by using the identities of sum and product of polynomials (the last one was found here: https://math.stackexchange.com/q/1937571).
Ok, now comes the question: If $F$ is, now, a domain, the result still true? Because in my proof I can't see any necessity of it been an field. If someone have some clue, I'll aprecciate!
You can prove the following more general result:
Lemma Let $R$ be a commutative ring with unity. Then $f: R[x] \to R[x]$ defined by $f(p)=P(aX)$ is an isomomorphism if and only if $a$ is invertible in $R$.
For the direct implication you can show that $g(P)=P(a^{-1}X)$ is the inverse.
For the inverse implication, use the fact that there exists some $P \in R[X]$ such that $$f(P)=X$$