About proving that $Aut(\mathbb{Z}_n)\simeq \mathbb{Z}_n^\times$. I need to prove that
$$
Aut(\mathbb{Z}_n) \simeq \mathbb{Z}_n^\times.
$$
My definition of $\mathbb{Z}_n$ is that
$$
\mathbb{Z}_n =\{\bar{m}: m\in \mathbb{Z}\}
$$
where $\bar{m}$ is the equivalence class containing $m$. I have defined the map
$$
\Psi: Aut(\mathbb{Z}_n) \to \mathbb{Z}_n^{\times}
$$
by
$$
\Psi(f) = f(\bar{1}).
$$
I have two questions about this
(1) Does this map even make sense? I mean, $f(1)$ is an element of the group $\mathbb{Z}_n$ and not $\mathbb{Z}_n^\times$. My thinking is that this is not a problem because $\mathbb{Z}_n^\times\subseteq \mathbb{Z}_n$. Is that right?
(2) Assuming that the map is well-defined, I am trying to show that this is a homomorphism. So I need $\Psi(fg) = \Psi(f)\Psi(g)$. I have
$$
\Psi(fg) = fg(\bar{1}) = f(g(\bar{1})).
$$
and
$$
\Psi(f)\Psi(g) = f(\bar{1})g(\bar{1}).
$$
I can see that on can thinking about $g(\bar{1})$ as just an integer and then it makes sense. But $g(\bar{1})$ is really an equivalence class. So how can I do this precisely?
 A: I think you can solve this problem much easier simply by switching the range and domain of your map. More precisely, define $f: \mathbb{Z^{\times}_n} \to Aut(\mathbb{Z_n})$ with $f(a):= f_a$; and define $f_a (x) = ax$ for all $x \in \mathbb{Z_n}$. 
A: I think you are refering to group automorphisms, rather than ring automorphisms.  Alternatively to Amin's idea, you can define $g:\text{Aut}\left(\mathbb{Z}_n\right)\to\mathbb{Z}_n^\times$ by $g(\varphi):=\varphi\left(\bar{1}\right)$ for all $\varphi\in \text{Aut}\left(\mathbb{Z}_n\right)$.
A: The definition of the group $\mathbb{Z}_n$ looks a bit strange to me. Maybe it is personal taste, but the equivalence relation is the essential part of the definition. So you should give the equivalence relation.
If you have a abelian group $G$ and a equivalence relation $\sim$ on $G$, such that the map
$$ G \to G/ \sim, g \mapsto [g] $$
which takes an element to its equivalence class, then you can show that $[0]$ is a Group and $[g]= g+ [0]$ for all $g \in G$. So you can define an equivalence relation
$$ a \sim_N b \Leftrightarrow  a-b \in N $$
for any subgroup $N$ of $G$. Then you can show that $G / \sim_N$ has a group structure by adding the elements of the equivalence classes. So it makes sense to denote this group by
$$ G/N $$
So giving the equivalence relation or the subgroup $n\mathbb{Z}$ would I expect in the definition of $\mathbb{Z}_n$.
To (1) a priori this map does not make sense. You can define the map with codomain $\mathbb{Z}_n$ and the show that it only takes values in $\mathbb{Z}_n^{\times}$.
For (2) you should look at generators as Nephry said. Any homomorphism between two groups is already defined on the generators. But you should be aware of the fact that it does not work the ofter way round. If you have generators and assign them elements in an other group, this does not always extend to a homomorphism. This only works in so called free (abelian) groups.
Now if you have a group $G$ with one generator $g$ you can show that any homomorphism $f \colon G \to G'$ is defined by $f(g)$.
In your case this means that you have to show that $\bar{1}$ is an automorphism if and only if $f(\bar{1})$ is a unit.
To calculate what $f(g(\bar{1}))$ is you need to express this by $f(1)$ which is possible since $\mathbb{Z}_n$ has one generator.
If you think of $g(\bar{1})$ as an integer you have to take care that it is independent of the representative you choose. You could also choose representatives for all classes to get rid of equivalence classes like $\mathbb{Z}_{n} \cong \{ 0,1, \dots, n-1\}$
Last but not least I prefer the notion $\mathbb{Z}/n \mathbb{Z}$, because by $\mathbb{Z}_n$ you also denote the localization of $\mathbb{Z}$ at the set $\{n^i \mid i\in \mathbb{N} \}$.
A: Let me try to address your questions and give the proof you are looking for. For your first question, the map $\Psi:\mathrm{Aut}(\mathbb{Z}_n)\to\mathbb{Z}_n^\times$ is correct, but you need to prove that $f(1)\in\mathbb{Z}_n^\times$ (this is not immediate).
Let's do this:
Suppose $f(1)=a$. Then, for $2\leq k<n$,
$$f(k)=f(1+\cdots+1)=\underbrace{f(1)+\cdots+f(1)}_k=ka=ak.$$
It follows that $f$ is multiplication by $a$ and we write $f=f_a$ in this case. To see that $a\in\mathbb{Z}_n^\times$, not that $f^{-1}=f_b$ for some $b\in\mathbb{Z}_n$ and
$$1=f^{-1}f_a(1)=f_bf_a(1)=f_b(a)=ba$$
Hence, $b=a^{-1}$ and $a\in\mathbb{Z}^{\times}$.
Now, we have seen that every $f\in\mathrm{Aut}(\mathbb{Z}_n)$ is of the form $f=f_a$ for some $a\in\mathbb{Z}_n^\times$, and it is clear that $f_a=f_b$ if, and only if $a=b$. This means, the map $\Psi$ above is surjective.
Now, to your last question, the previous computation also shows that $f_bf_a=f_{ba}$ for any $a,b\in\mathbb{Z}$. This means that
$$
\Psi(f_bf_a)=\Psi(f_{ba})=f_{ba}(1)=ba=f_b(1)f_a(1)=\Psi(f_b)\Psi(f_a)
$$
so $\Psi$ is a homomorphism.
It is interesting to point out that we are regarding $\mathrm{Aut}(\mathbb{Z}_n)=\mathrm{Aut}_{\mathrm{Group}}(\mathbb{Z}_n)$ as the group of Group automorphisms. You may also know that $\mathbb{Z}_n$ is a Ring. In this context, we have $\mathrm{Aut}_{\mathrm{Ring}}(\mathbb{Z}_n)=\{1\}$ since ring automorphism must additionally satisfy $\phi(1)=1$. This little piece of trivia is actually a preview of something you may come to understand as categorical considerations.
