Conformal equivalence and the automorphism group of a domain

Let $U,V\subseteq \mathbb{C}$ be complex domains. $U$ and $V$ are conformally equivalent when there exists a biholomorphism $f:U\to V$ in the sense of Biholomorphic maps between complex domains; such an $f$ is then a conformal equivalence from $U$ onto $V$.

For a complex domain $U$, the automorphism group of $U$ is

$\mathrm{Aut}(U):=\{f:U\to U:f\text{ is biholomorphic}\},$

with composition as the group operation.

Why the group operation is legitimate. The identity map ${\mathrm{id}}_U$ is biholomorphic. If $f$ is biholomorphic then its inverse $f^{-1}$ is holomorphic by the definition of biholomorphy, so $f^{-1}\in \mathrm{Aut}(U)$ when $f\in \mathrm{Aut}(U)$. If $f,g\in \mathrm{Aut}(U)$ then the composite $g\circ f$ is biholomorphic: it is a bijection whose inverse $f^{-1}\circ g^{-1}$ is a composite of holomorphic maps, hence holomorphic. Composition of maps is associative, so these three closure facts make $\mathrm{Aut}(U)$ a group with identity ${\mathrm{id}}_U$.