Find the group of conformal automorphisms of $U=\lbrace z\in \mathbb{C}: \vert z-1\vert>1\rbrace$ Well $\phi$ is an automorphism of $U$  $\iff$ $1/ \phi$ is an automorphism of $U^C=\lbrace z\in \mathbb{C}:\vert z-1\vert<1\rbrace$ $\iff$ $1/\phi -1$ is an automorphism of the unit disc $\iff$ $1/\phi(z) -1=\omega \frac{z-a}{1-\bar a z}$ for some $w\in \mathbb{C}$ with $\vert \omega \vert=1$ and for some $a\in \mathbb{C}$ with $0\le \vert a\vert<1$. Hence $\phi(z)=\frac{1}{1+\omega\frac{z-a}{1-\bar az}}.$ So $Aut(U)$ is the group consisting of all maps of the form $$\frac{1}{1+\omega\frac{z-a}{1-\bar az}}$$ where $\vert \omega \vert =1$ and $0\le \vert a\vert <1$. Is my work correct, or is there a more explicit description of the group? 
Additional practice: 
I would like additional practice with this type of problem, are there any guidlines I should have when choosing a region $U$ to start with - or does anyone have any additional examples in mind? Please include answers to the latter as an answer if you have one.
 A: Your $\Leftrightarrow$s are incorrect - I'll give you the correct version.
An important idea to put into play here is that conjugation changes perspective. For instance, one conjugates a matrix by a change-of-basis matrix in order to obtain the matrix of the same linear transformation with respect to a new basis. As a second example, conjugating a permutation by a second relabels the entries in its disjoint cycle representation according to the action of the second permutation. Third example, if a group $G$ acts on a set $X$, then ${\rm Stab}(gx)=g{\rm Stab}(x)g^{-1}$ for any $g\in G$ and $x\in X$. We see something similar in this situation:

Suppose $\tau:U\to V$ is an isomorphism in some category. Then $$\phi\in{\rm Aut}(U)\iff \tau\circ\phi\circ\tau^{-1}\in{\rm Aut}(V). \tag{$\circ$}$$ In particular we may consider a category whose objects are open subsets of $\Bbb C$ and whose morphisms are holomorphic maps. (Don't worry if you don't know what a category is.)

One way to understand why this is true is to consider the following commutative diagram:
$$\begin{array}{ccc} U & \to & V \\ \downarrow & & \downarrow \\ U & \to & V \end{array} $$
The horizontal maps are both $\tau:U\to V$ for some isomorphism $\tau$. Any automorphism of $U$ (the first vertical map down) can be obtained from an automorphism of $V$ by simply following along the long way from $U$ to $V$ in the above diagram. Why? Well, because the process is reversible: the inverse of going from $\phi$ to $\tau\circ\phi\circ\tau^{-1}$ is going from $\psi$ to $\tau^{-1}\circ\psi\circ\tau$. (This will be a lot clearer if you have some familiarity with basic group theory.)
In order to simplify the whole situation with linear fractional transformations, we can use the convenient matrix notation $(\begin{smallmatrix}a&b \\ c&d\end{smallmatrix})z:=\frac{az+b}{cz+d}$. I leave it to you to check that $(AB)z=A(Bz)$ are identical functions of $z$ for any matrices $A,B\in{\rm GL}_2(\Bbb C)$. This defines a group action of matrices on the complex plane (or really the Riemann sphere), if you're familiar with the concept.
Suppose $|z-1|>1$. This is the unit circle centered at $1$. If we translate $1$ unit to the left and then reciprocate (notice the order of operations, and the fact we're visualizing the situation in order to see what's going on) we wind up with the unit disk $\Bbb D$ (assuming complex infinity was an option for $z$, which corresponds to $0\in\Bbb D$). Thus $\tau(z)=\frac{1}{z-1}=(\begin{smallmatrix}0&1 \\ 1&-1\end{smallmatrix})z$ defines an isomorphism between the open sets $U$ and $\Bbb D$.
There is a classification of the maps in ${\rm Aut}(\Bbb D)$. They are the Blaschke factors composed with rotations, in other words $\phi(z)=(\begin{smallmatrix}\omega & -\omega a \\ -\bar{a} & 1\end{smallmatrix})z$ for some $a\in\Bbb D$ and $\omega\in S^1$.
Therefore, by $(\circ)$, every automorphism of $U$ looks like $(\begin{smallmatrix}0&1 \\ 1&-1\end{smallmatrix}) (\begin{smallmatrix}\omega & -\omega a \\ -\bar{a} & 1\end{smallmatrix}) (\begin{smallmatrix}0&1 \\ 1&-1\end{smallmatrix})^{-1}$ for appropriate parameters $a$ and $\omega$. I leave it to you to perform the matrix operations in order to simplify.
Here's what you want to do now. Figure out a procedure for constructing an isomorphism between any two generalized disks $U$ and $V$. (By generalized disk, I mean a standard open disk, the interior of the complement of one, or a half-plane. In other words, the possible images of the unit disk under a linear fractional transformation.) In fact this problem is equivalent to merely determining an isomorphism $U\to\Bbb D$ (do you see why?). Remember, the basic operations are translation, dilation, and inversion, and you need to do them in the right order to succeed.
If you ever see another problem like this again with relating ${\rm Aut}(U)$ and ${\rm Aut}(V)$ where $U,V$ are two generalized disks, you can find an isomorphism $\tau:U\to V$ and then use principle $(\circ)$ in order to complete the problem.
More generally, the Riemann mapping theorem says that any simply connected open subset of $\Bbb C$ is isomorphic to the unit disk $\Bbb D$, so this task could in general be done with any two such regions $U$ and $V$ of the plane, but the general problem of determining a holomorphic bijection $U\to \Bbb D$ is essentially solving a two-dimensional boundary value problem from partial differential equations using the so-called Dirichlet principle, and this is too far afield do discuss on a tangent.
Just a little more tangent: the genus $1$ regions in the plane (i.e. open subsets with a single hole in them) are all isomorphic to a unique annulus with inner radius $1$. I don't recall if there is any known classification of regions with genus $2$ (i.e. two holes).
