Prove that 2 groups $Aut(D),Aut(D')$ are isomorphism Prove that two groups $\operatorname{Aut}(D),\operatorname{Aut}(D')$ are isomorphic given that there is a bijective and holomorphic function $f$ from region $D$ onto region $D'$ (in other words, $f^{-1}$ is also holomorphic).
Any idea how to show this? I have to find a mapping $g\colon\operatorname{Aut}(D)\to\operatorname{Aut}(D')$ such that $g(xy)=g(x)g(y)$ for all $x,y$ and $g$ also 1-1 and onto, but I have no idea now.
 A: First, the correct terminology is to show that the two groups are isomorphic. A way to do that is to exhibit an isomorphism between the groups. So, we wish to construct a function $g:Aut(D)\to Aut(D')$, and then show that it is a homomorphism and invertible (since that will show that it is an isomorphism of groups). So, given an element $x$ in $Aut(D)$, what can possibly $g(x)$ be? Well, $x$ is actually a function $x:D\to D$, and we wish $g(x)$ to be a function $g(x):D'\to D'$. With the information given, there is a clear candidate for such a construction: $g(x)=f\circ x \circ f^{-1}$. You can now check this this definition indeed gives a homomorphism and that it is a bijection (hint for a very quick proof that it is a bijection: find the inverse).
The geometric meaning of this exercise should not escape attention: If the two regions are 'the same', in the sense that there is a structure preserving bijection between the two, then the groups of isometries of each region (which here can be taken to be the set of all holomorphic bijections of the region) are themselves isometric. In short, if two shmoofs are essentially identical, then their shpeefs of symmetries are essentially identical. 
