Can someone explain automorphisms to me? So I know the definition of an automorphism is an isomorphism that maps from a group to itself. How can an element of an automorphism map to something besides itself?
 A: Consider the group of rotations of a square.  This is isomorphic to the cyclic group $C_4$, generated by rotating the square $90^\circ$ clockwise.  If I perform this rotation four times, I get back to where I started.  
But I could also start by rotating the square $270^\circ$ clockwise - that is, rotating it $90^\circ$ anti-clockwise.  The group then has precisely the same structure: this gives us an automorphism: 
\begin{align}
0^\circ\textrm{ clockwise }&\mapsto0^\circ\textrm{ clockwise }\\
90^\circ\textrm{ clockwise }&\mapsto270^\circ\textrm{ clockwise }\\
180^\circ\textrm{ clockwise }&\mapsto180^\circ\textrm{ clockwise }\\
270^\circ\textrm{ clockwise }&\mapsto90^\circ\textrm{ clockwise }\\
\end{align}
This is a pretty simple example.  Some more interesting ones are: 


*

*Complex conjugation - the map $a+bi\mapsto a-bi$ is an isomorphism of both $(\mathbb C, +)$ and $(\mathbb C\setminus\{0\},\times)$

*Automorphisms of $(\mathbb Q, +)$: these are bijective functions $f$ on $\mathbb Q$ satisfying $f(x+y)=f(x)+f(y)$, and turn out to be functions of the form $f(x)=kx$ for $k\ne0$.  


The automorphisms of a group always form a group themselves (Why?) : for instance, the automorphism group of $(\mathbb Q, +)$ is isomorphic to $(\mathbb Q\setminus\{0\},\times)$ (Why?)
A: Here are some other examples that make for nice exercises:
Let $G = C_3$ be the cyclic group of order $3$. Let $\varphi: G\to G$ be the map that fixes the identity and switches the other two elements. Show that this is an automorphism of $G$.
Let $G = C_2\times C_2$ be the Klein four-group. Let $\varphi: G\to G$ be any bijective map fixing the identity. Show that $\varphi$ is an automorphism of $G$.
A: The automorphisms give you an idea of the symmetries of the group.
Consider the automorphism $\phi : \mathbb{C} \to \mathbb{C}$ given by $\phi(z) = \overline{z}$, i.e. $\phi(x+\operatorname{i}\!y) = x - \operatorname{i}\!y$. This is an automorphism and $\phi$ only fixed $z=0$.
Consider the extended complex plane $\widehat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$, also called the Riemann Sphere. The automorphisms here are of the form
$$\phi(z)=\frac{az+b}{cz+d}$$
where $a,b,c,d \in \mathbb{C}$ and $ad-bc \neq 0$.
A: Firstly, an automorphism is a function from $G$ to $G$. As the other answers indicate, it can certainly take a group element to something other than itself.
To answer your question, if $f : G\to G$ given by $f(x) = x^2$ is an automorphism, then at the very least, it is a homomorphism. So
$$
(xy)^2 = f(xy) = f(x)f(y) = x^2y^2
$$
So
$$
xyxy = xxyy
$$
Now what happens when you cancel on both sides?

Furthermore, $f$ is injective, and so $\ker(f) = \{e\}$, and so there is no $x\neq e$ such that $x^2 = e$. In other words, $G$ does not have any element of order 2. By Cauchy's theorem, this implies that $|G|$ is not a multiple of 2.
Once $f$ is injective, it is automatically surjective (why?), so this is all you can say about $G$.
