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 $Z\subset F_g$. I claim that
$F_g$ is a convex subset of $T$. Indeed, take two points $p, q\in F_g$. Then the Teichmuller geodesic segment $pq$ between $p, q$ is mapped via $g$ to a path $c$ in $T$ of length $\le d(p,q)$ (since $g: T\to T$ is a holomorphic map, it weakly decreases the Kobayashi-Teichmuller distance). But $c$ connects $p$ and $q$, hence, its length $=d(p,q)$, i.e. $c$ is a geodesic segment equal to $pq$; thus, $pq$ has to be fixed by $g$ pointwise. Thus, $pq\subset F_g$.
(Most likely, $F_g$ is a submanifold but we do not need this.) Convexity of $F_g$ implies that 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 and locally contractible finite-dimensional locally compact metrizable space (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 and locally contractible subset, 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.
