Clever way of showing the map sending to prime power is outer automorphism

Question

I was asked to find an outer automorphism of a finite group and came up with $x\mapsto x^3$ for $Z_4$ (cyclic group order 4). I went through a lengthy check of all properties to prove that this was indeed an outer automorphism when it occurred to me there probably exists a slicker way to show this. Is there?

Answer 1

The only inner automorphism of an abelian group is the identity map!

Answer 2

Let $a$ be a generator of a cyclic group $C_n$ of order $n$. An automorphism $\alpha$ (one can drop the adjective outer for abelian groups as made clear in the comments) of $C_n$ is entirely known if its image $\alpha(a)$ is known. Indeed, the image of an arbitrary element $a^i \in C_n$ is then $\alpha(a)^i$. So an automorphism of $C_n$ is uniquely determined by a number $d$, namely by the condition $\alpha(a) = a^d$. The inverse is not true. Let $\gcd(d,n) = k > 1$. Let $k'$ be such that $kk'=n$. The element $a^{k'}$ is not $1$ since $k' < n$, but $\alpha(a^{k'}) = 1$ since $dk'$ is a multiple of $n$. The only automorphisms are defined by those $d$ such that $\gcd(n,d) = 1$.