Let $A$ be an integral domain, and $c\in A$. Let $h:A[x]\rightarrow A[x]$ be defined by $h(a(x))=a(cx)$, that is, $$h(a_0+a_1x+\ldots+a_nx^n)=a_0+a_1cx+\ldots+a_nc^nx^n.$$ Prove that $h$ is a homomorphism and describe its kernel. Prove that $h$ is an automorphism if and only if $c$ is invertible.
To show that $h$ is a homomorphism, we note that $$h(a(x)+b(x))=(a+b)(cx)=a(cx)+b(cx)=h(a(x))+h(b(x)).$$
Also, $$h(a(x)b(x))=(a\cdot b)(cx)=a(cx)b(cx)=h(a(x))h(b(x)).$$
$a(x)$ is in the kernel iff $a_0=a_1c=\ldots=a_nc^n=0$. (Is there an easier way to describe this?)
Now, I'm not sure what to do with the automorphism (which means checking surjectivity and injectivity).
Because it is an integral domain, if $a_ic^k=0$, either $a_i=0$ or $c^k=0$.
Let $r$ be the minimum power $k$ for which $c^k=0$. Then we get that $a_i=0$ for $i<r$ is the condition for $a_0+a_1x+...a_nx^n$ being in the kernel.
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If $c$ is a unit then $g(p(x)):=p(c^{-1}x)$ is the inverse of $h$.
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Assume now that $h$ is invertible with inverse $g$. Then $a_0=g(h(a_0))=g(a_0)$. So, $g$ fixes coefficients (the elements of $A$). Because of this $g(ap(x))=g(a)g(p(x))=ag(p(x))$, for $a\in A$.
Also $g(h(x))=g(cx)=cg(x)=x$. Therefore $g(x)=a_nx^n...+a_1x+a_0$, with $ca_k=0$, for $k\neq1$, and $ca_1=1$. If $c=0$ then $h(p(x))=0$, which is not invertible. Therefore $c$ is invertible with inverse $a_1$, and $a_k=0$, for $k\neq1$.