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 where is the open unit disc. Consider given by . Now is a conformal map from the unit disc to itself, and therefore has the form for some . If , then the fixed points of this transformation are the two solutions of , which lie on the boundary of (and hence they are not fixed points of , whose domain does not include the boundary). So the only possibility is , i.e. and .