The statement is false as written; what about a constant map? I suppose you are assuming the map to be conformal; if so, then it's true and may be proven like this :

By Riemann's mapping theorem, we have a conformal map \(\displaystyle \sigma : G \rightarrow D\) where \(\displaystyle D\) is the open unit disc. Consider \(\displaystyle F : D\rightarrow D\) given by \(\displaystyle F(z) = (\sigma f\sigma^{-1})(z)\). Now \(\displaystyle F\) is a conformal map from the unit disc to itself, and therefore has the form \(\displaystyle F(z) = \frac{z-a}{1-\overline{a}z}\) for some \(\displaystyle |a|<1\). If \(\displaystyle |a|>0\), then the fixed points of this transformation are the two solutions of \(\displaystyle z^2=\frac{a}{\overline{a}}\), which lie on the boundary of \(\displaystyle D\) (and hence they are *not* fixed points of \(\displaystyle F\), whose domain does not include the boundary). So the only possibility is \(\displaystyle |a|=0\), i.e. \(\displaystyle F(z)=z\) and \(\displaystyle f(z)=z\).