Order of automorphism group of cyclic group Let $G$ be a cyclic group of order $m$. What is the order of $\text{Aut}(G)$?
I want to know the proof as well (elementary if possible). I would still accept the proof if one answers with $m = p$, a prime. Or on top of that, I would accept the answer with extra assumption: $q \equiv 1$ mod $p$ with another prime $p$.
 A: Since an automorphism of $G$ should map a generator of $G$ to a generator of $G$ it's enough to know how many generators does $G$ have. 
If $G=\{e,g,g^2,...,g^{m-1}\}$ then a $g^i$ generates G if and only if $\operatorname{gcd}(i,m)=1$. 
$\lvert \operatorname{Aut}(G)\rvert=\phi(m)$ where $\phi(m)$ is Euler's function. 
For a more detailed proof:


*

*Let $G=\langle g\rangle$ and $f\in\operatorname{Aut}(G)$. 

*Then $f(g)=g^i$ for some $i$. If $f$ is an isomorphism  $\langle g^i\rangle =G$ and this happens only if $\operatorname{gcd}(i,m)=1$.

*On the other hand every homomorphism $f:G\rightarrow G$ with $f(g)=g^i$ is an isomorphism when $\operatorname{gcd}(i,m)=1$, so $\lvert \operatorname{Aut}(G)\rvert=\phi(m)$.

A: Hint: 
For a given cyclic group $G,\;\text{with}\;\; |G| = m$:
$$\text{Aut}\,(G) \cong \mathbb{Z}_m^*\tag{$\;^*:\;\;$multiplicative group}$$
Hence $\text|{Aut}\,(G)| = |\mathbb{Z}_m^*|.$
Since an automorphism of $G$ should map a generator of $G$ to a generator of $G$, it's enough to know how many generators $G$ must have.
So $|\text{Aut}(G)|=|\mathbb{Z}_m^*| = \phi(m)$ where $\phi(m)$ is Euler's totient function. 

(Note: you do not need that $m = p$, a prime; it suffices to know that $G$ is cyclic and finite.)
