Let $T$ denote the Teichmuller space of a hyperbolic Riemann surface of finite conformal type. Suppose that $f: T\to T$ is a holomorphic map which has a periodic orbit $Z$, i.e. a finite invariant subset on which $f$ acts as an order $n$ cyclic permutation. Set $g=f^n$, then $g$ fixes $Z$ pointwise. Since $f$ and $g$ commute, $f(F_g)\subset F_g$, where $F_g$ is the fixed-point set of $g$ in $T$. It is nonempty since $Z\subset F_g$. The subset $F_g$ is a convex complex submanifold of $T$. (I did not check carefully that it is a submanifold, a priori, it is only an analytic subvariety. Convexity should imply that it is a submanifold. In the worst case, a subvariety suffices for the proof, since all I need is that it admits a triangulation.) In particular, it is contractible. The restriction of $f$ to $F_g$ is a periodic homeomorphism of order $n$. It is now a classical argument going back to Nielsen that $f$ has a fixed point in $F_g$. It goes as follows. Set $F:=F_g$ and 
$h:=f|_F$. Suppose first that $n$ is prime. A finite cyclic group of prime order cannot act freely on a contractible finite-dimensional CW complex (since finite nontrivial groups have infinite cohomological dimension). Hence, some nontrivial element of $\langle h\rangle$ fixes a point in $F$. But, since $n$ is prime, this element generates the cyclic group  $\langle h\rangle$. The general case is proven by induction on the number of prime factors of $n$. Suppose that $n=pq$ where $p$ is prime. Then $h^q$ has order $p$ and, hence, has nonempty fixed-point set $F'=F_{h^q}$ in $F$. But $F'$ is again convex, hence, a contractible submanifold/subvariety, invariant under the action of $h$. Thus, we reduced the  problem to finding a fixed point of $h^p$ in $F'$. The order of $h^p$ is $q<n$ and we continue inductively.