Let $G=\langle a\rangle$ be a cyclic group.
- Let $\eta\in\mathrm{End}\langle a\rangle$. Then "$\eta\in\mathrm{Aut}\langle a\rangle\Leftrightarrow \eta(a)$ is a generator of $\langle a\rangle$".
- Proof. If $\left |\langle a\rangle\right |=\infty$, $\mathrm{Aut}\langle a\rangle=\{\mathrm{id}\}$. Suppose $\left |\langle a\rangle\right |=n<\infty$. ($\Rightarrow$) $\eta (a^k)=\left(\eta (a)\right)^k$ and $\eta$ is an epimorphism. ($\Leftarrow$) If $1=\eta (a^k)=\left(\eta (a)\right)^k$ then $n\bigl | k$.
- Let $\left |\langle a\rangle\right |=n$. Then $\mathrm{Aut}\langle a\rangle=\left\{\eta\in\mathrm{End}\,\left |\,\eta(a)=a^k,\,(n,k)=1\right\}\right.$
- Proof. Corollary of the above. (Note: There exists $\alpha,\beta\in\mathbb{Z}$ such that $\alpha n+\beta k=1$. Hence, $\left(a^{k}\right)^\beta=a$.)
- Let $\left |\langle a\rangle\right |=n$ and $U\left(\mathbb{Z}/(n)\right)$ be the multiple group of $\mathbb{Z}/(n)$. Then \begin{align*}\begin{array}{ccc}\mathrm{Aut}\langle a\rangle&\cong&U\left(\mathbb{Z}/(n)\right)\\\eta_k&\mapsto&\bar{k}=k+(n)\end{array}\end{align*} where $\eta_k(a)=a^k$, $(n,k)=1$.
