What is $\mbox{Aut}\left(\mathbb{Z}_{n}\right)$? I understand that $\mbox{Aut}\left(\mathbb{Z}_{n}\right)$ is the group of all automorphisms under function composition, but I am a little confused about the sort of group it forms. If the elements of this group are homomorphisms, then how is $\mbox{Aut}\left(\mathbb{Z}_{n}\right)$ isomorphic to $\mbox{U}(n)$? Doesn't $\mbox{U}(n)$ contain numbers as elements, such that $(x,n)=1$, where $x \in \mbox{U}(n)$? As an example, can someone let me know what the elements would be in $\mbox{Aut}\left(\mathbb{Z}_{10}\right)$?
 A: First of all, it seems you are confusing isomorphism and equality. The elements of the automorphism group are indeed functions, but they correspond to the invertible elements $\mathbb Z/n\mathbb Z^*$ in a very specific way which also preserves the group structures.
Now, why is $Aut(\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z^*$? Recall that a morphism from $\mathbb Z/n\mathbb Z$ is completely determined by where it sends $1$, since $k$ is simply the sum of $k$ copies of $1$. Therefore, for a morphism 
$$f:\mathbb Z/n\mathbb Z\to\mathbb Z/n\mathbb Z,$$
there are $n$ possibilities for $f(1)$. But $f$ will be invertible if and only if $f(1)$ is coprime to $n$ (why?).

For example, if we look at the group $\mathbb Z/10\mathbb Z$, the automorphisms are given by
\begin{align*}
f_k:\mathbb Z/10\mathbb Z &\to \mathbb Z/10\mathbb Z\\
a &\mapsto ka,
\end{align*}
where $\gcd(k,10)=1$. Since the integers coprime to 10 have residues modulo $10$ equal to $1,3,7,9$, we conclude that 
$$Aut(\mathbb Z/10\mathbb Z)=\{f_1,f_3,f_7,f_9\}.$$
A: I assume you are regarding $\mathbb Z_n$ as an additive abelian group.
The elements of $\mathbb Z_{10}$ are then $0,1,2,3,4,5,6,7,8,9$. Suppose we multiply them all by $3$ to get $0,3,6,9,2,5,8,1,4,7$ (we work modulo $10$). Note that $3a+3b=3(a+b)$ so we have a homomorphism. And the explicit list shows that the homomorphism is an isomorphism (kernel is trivial: inverse is multiply by $7$ if you need to check).
If we multiply by $4$ we still have a homomorphism, but the image is $0,4,8,2,6,0,4,8,2,6$, and if we try $5$ we get $0,5,0,5,0,5,0,5,0,5$.
That should be enough to illustrate what is going on here.
