I want to show that $\text{Aut}(\mathbb{Z}_n)$ is an abelian group of order $\phi (n)$.
The automorphism group of the group $\mathbb{Z}_n$, $\text{Aut}(\mathbb{Z}_n)$, is the group of isomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_n$.
How do the elements of $\text{Aut}(\mathbb{Z}_n)$ look like?
Suppose $\phi\in\mathrm{Aut}(\mathbb{Z}_n)$ and suppose $\phi([1])=[m]$. Then, for $0\leq r<n$ we have \begin{align*} \phi([r])=&\phi([\underbrace{1+\cdots+1}_r])\\ &=\phi(\underbrace{[1]+\cdots+[1]}_r)\\ &=\underbrace{\phi([1])+\cdots+\phi([1])}_r=\underbrace{[m]+\cdots+[m]}_r=[mr] \end{align*} Hence, $\phi$ is multiplication by $m$ and we may write $\phi=\phi_m$. But, $\mathrm{Aut}(\mathbb{Z}_n)$ is closed under inverses and we have $\phi_m^{-1}=\phi_k$ for some $k$. Now, $$ [1]=\phi_m^{-1}\phi_m([1])=\phi_k\phi_m([1])=\phi_k([m])=[mk]=[m][k] $$ Hence, $[k]=[m]^{-1}$ and $[m]\in\mathbb{Z}_n^{\times}$.
Now, a computation similar to the one above also shows that $\phi_r\phi_s=\phi_{rs}$. Hence, the map $$\Phi:\mathrm{Aut}(\mathbb{Z}_n)\to\mathbb{Z}_n^{\times}$$ given by $\Phi(\phi_m)=[m]$ is a homomorphism. It is now straighforward to check that this map is bijective. Hence, $\mathrm{Aut}(\mathbb{Z}_n)$ is an abelian group of order $|\mathbb{Z}_n^\times|=\varphi(n)$.