$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.

Let $G=\langle a\rangle=\mathbb Z_n$ and get $\phi\in Aut(G)$. Clearly, $$\phi(a)=ta:=\underbrace{a+a+\ldots+a}_t$$ for some $t$. You know that $ta$ is a generator of the group and therefore $(t,n)=1$ necessarily. Here you have $[t]\in U(\mathbb Z_n)$. Now try to show that the following function is an isomorphism: $$\Phi: Aut(G)\longrightarrow U(\mathbb Z_n)$$ $$\Phi(\phi)=[t]$$

This is a fleshed out version of Brian's hint.

Lemma Let $1\in \mathbb{Z}_n$. If $\varphi\in \operatorname{Aut} \mathbb Z_n$, then $\varphi(1)$ must be a generator of $\mathbb Z_n.$

Proof: Because $\varphi$ is an automorphism (of an abelian group), $\varphi(kx)=k\varphi(x)$ for $k\in \mathbb Z, x\in \mathbb Z_n$. Additionally, $\varphi(0)=0$. Therefore if $k\varphi(1)=0$ for some $k<n$, then applying $\varphi^{-1}$ to both sides yields $k1=0$

By the lemma, we must have $\varphi(1)$ is a generator of $\mathbb Z_n$. The generators are the elements relatively prime to $n$ when the elements of $\mathbb Z_n$ are viewed as a subset of $\mathbb Z$, which are in correspondence with the elements of $U_n$. Moreover, $\varphi(\ell)=\varphi(\ell 1)\equiv \ell\phi(1) \pmod n$, and so the automorphism is given by multiplication by $\varphi(1) \mod n$, and so the map $\Psi:\operatorname{Aut}\mathbb Z_n\to U_n$ sending $\varphi$ to multiplication by $\phi(1)$ is injective. It is not hard to check that it is also surjective. Let us show that it is a group homomorphism.

Let $\varphi,\psi\in \operatorname{Aut}\mathbb Z_n$. Then $(\varphi \circ \psi)(1)=\varphi(\psi(1))=\varphi(1)\psi(1)$ by our calculation above. Therefore $\Psi(\varphi \circ \psi)=\Psi(\varphi)\Psi(\psi)$

HINT: Suppose that $\varphi\in\operatorname{Aut}\Bbb Z_n$; then $\varphi(1)\in U_n$. (Why?) Consider the map $$h:\operatorname{Aut}\Bbb Z_n\to U_n:\varphi\mapsto\varphi(1)\;.$$