Let $U\subset C$ be a simply connected domain, $U \neq C$ and $z_0 \in U$. I want to prove that
$\phi: \{f: U\rightarrow U$ biholomorphic, $f(z_0) = z_0 \} \rightarrow C, f \mapsto f'(z_0)$
defines an isomorphism of groups $Aut_{z_0}(U) \cong S^1$. It is clear for me that $f$ is a homomorphism, but I do not have an idea how to show that $im f \subset S^1$ and that $f$ maps bijectively onto $S^1$. There is also a hint in the task that one could use Schwarz' Lemma.
Glougloubarbaki has given the crucial ingredients: Riemann's mapping theorem and Schwarz's lemma.
Consider $D = \lbrace z \in \mathbb{C} \mid \lvert z \rvert < 1 \rbrace$ and a biholomorphic $f : D \to D$ such that $f(0) = 0$. By Schwarz's Lemma $\lvert f'(0) \rvert \le 1$ and $\lvert (f^{-1})'(0) \rvert \le 1$. Since $(f^{-1})'(0)f'(0) = 1$ (chain rule), we get $\lvert f'(0) \rvert = 1$. Again by Schwarz's lemma we obtain $f(z) = a z$ for some $a \in S^1$. But then $a = f'(0)$. Since all $f_a(z) = a z$, $a \in S^1$, are biholomorphic and $f_a = f_b$ iff $a = b$, we are done.