Automorphic group of a cyclic group

Question

Suppose $G$ is a cyclic group of order $n.$ Then $Aut(G)\cong (\mathbb{Z}/n\mathbb{Z})^\times$. Why is this true?

Answer 1

Hint: $f:G\to G$ by $f(a)=a^k$ is an isomorphism if and only if $(k,n)=1$ where $G$ is a cylic group of order $n$.

First try to show above lemma after that try to show the converse i.e any isomorphism correspond to a such $f$ for some $k$.

Answer 2

let $a$ be a generator for $G$. any automorphism $\gamma$ must map $a$ to $a^g$ where $g$ is relatively prime to $n$ (an automorphism must preserve orders of elements). the rest follows

Answer 3

Hint: First use the face that $\varphi:\, a\mapsto a^{k}$ is an automorphism of a cyclic group iff $(k,n)=1$.

Then, for a general group $G$, $$ \varphi(\langle x\rangle)=\langle x\rangle\iff(k,n)=1 $$

and note that $$ G=\cup_{x\in G}\langle x\rangle $$