Automorphism group of a cyclic group

Statement

Let $n$ be a positive integer and let $G$ be a cyclic group of order $n$. For each integer $a$ with $\gcd(a, n) = 1$, the map $\sigma_a \colon x \mapsto x^a$ is an automorphism of $G$. The map $a \bmod n \mapsto \sigma_a$ is an isomorphism from the group of units $(\mathbb{Z}/n\mathbb{Z})^\times$ to $\operatorname{Aut}(G)$.

Remarks

An automorphism of $G = \langle g \rangle$ is determined by the image of $g$, and this image must be a generator $g^a$ with $\gcd(a, n) = 1$. Thus $|\operatorname{Aut}(G)| = \varphi(n)$, where $\varphi$ is Euler's totient function. For the infinite cyclic group, $\operatorname{Aut}(\mathbb{Z}) \cong \{1, -1\}$.

Explanations

L1 human

An automorphism of a cyclic group must send a generator to a generator. Each choice of a generator gives exactly one automorphism. For $\mathbb{Z}/5\mathbb{Z}$, the generators are $1, 2, 3, 4$, so there are four automorphisms, given by multiplication by $1, 2, 3, 4$.