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
.