I want to prove $(\mathbb{Z}/n\mathbb{Z})^{\times}\cong \text{Aut}(\mathbb{Z}_n)$ so
(1) I proved $f_s:G\to G $ such that $g\mapsto sg=g+\overset{s}{\dots}+g\,$ is an automorphism on $G $. ($\gcd(s,|G|)=1)$
(2) I proved $f:\mathbb{Z}_n\to\mathbb{Z}_n$ is an automorphis on $\mathbb{Z}_n \Leftrightarrow f([1])\in (\mathbb{Z}/n\mathbb{Z})^{\times}$
Finally I made $\phi:(\mathbb{Z}/n\mathbb{Z})^{\times}\to\text{Aut}(\mathbb{Z}_n)$ such that $[x]\mapsto \phi_{[x]}([t])=x[t]$ so I must prove $\phi $ is an isomorphism.
The $\gcd(|G|,x)=1$ because $x\in(\mathbb{Z}/n\mathbb{Z})^{\times}$ and by (1) $\text{Im}\,\phi\in \text{Aut}(\mathbb{Z}_n)$
Then I proved $\phi $ is an homomorphism, let $g,h\in(\mathbb{Z}/n\mathbb{Z})^{\times}:$ $$\phi_{[g][h]}([t])=(\phi_{[g]}\circ\phi_{[h]})([t])$$
Now I should prove $\phi $ is injective, for $g,h\in (\mathbb{Z}/n\mathbb{Z})^{\times}$ such that $\phi_{[g]}([t])=\phi_{[h]}([t])$ $$g[t]=h[t]$$ $$\vdots$$ $$g=t?$$
I am not sure how prove it ..I think maybe if I multiply both sides by $[t^{-1}]$ I could get the result?
$\phi$ is surjective because if $\varphi\in \text{Aut}(\mathbb{Z}_n)$ then by (2) exists $g=\varphi([1])\in (\mathbb{Z}/n\mathbb{Z})^{\times}$ such that $\phi_g=\varphi $ ?
Also I should prove $\phi$ is well-defined?
Any hint please?
Thank you for your help
Every automorphism of the additive cyclic group ${\mathbb Z}_n$ must map $1$ to a generator $g$ of that group. It is easy to see that $g$ is a generator iff $GCD(g,n)=1$. So there is a map from ${\rm Aut}({\mathbb Z}_n)$ to ${\mathbb Z}_n^+$. On the other hand, every choice of $g$ which is coprime with $n$ gives a map $\phi$:$[k]\mapsto k[g]$ of ${\mathbb Z}_n$ to itself. It is easy to see that $\phi$ respects the operation and is a bijection. Hence $\phi$ is an isomorphism.