Automorphisms of a cyclic group

Proposition (Automorphism group of a finite cyclic group). Let $G$ be a cyclic group of order $n$, and identify $G\cong \mathbb Z/n\mathbb Z$ via the standard isomorphism . Then

where $(\mathbb Z/n\mathbb Z)^\times$ denotes the group of units of the ring $\mathbb Z/n\mathbb Z$.

Equivalently: if $G=\langle g\rangle$, then every automorphism $\alpha\in \mathrm{Aut}(G)$ is uniquely determined by $\alpha(g)=g^k$ with $\gcd(k,n)=1$, and composition corresponds to multiplication of $k$ modulo $n$.

Context. This makes automorphisms of cyclic groups completely explicit: an automorphism is exactly the choice of a generator-image. The group $\mathrm{Aut}(G)$ itself is a central object in extension theory and semidirect products.

Proof sketch. Define a map $(\mathbb Z/n\mathbb Z)^\times\to \mathrm{Aut}(\mathbb Z/n\mathbb Z)$ by sending $\overline{k}$ to multiplication-by-$\overline{k}$. This is a homomorphism. It is injective because multiplication-by-$\overline{k}$ is the identity only when $\overline{k}=\overline{1}$. It is surjective because any group endomorphism of a cyclic group is determined by the image of $1$, and it is invertible exactly when that image is a generator, i.e. a unit modulo $n$.