I would like to show the following: Let $X$ be a complete metric space that is uniquely geodesic (i.e. each two distinct points are connected by a unique geodesic segment) and $\phi\colon X\to X$ an isometry of finite order $k$. Then $\phi$ has a fixed point.
My ideas so far:
Pretty elementary: The statement is obvious if $k\leq 2$. So take an orbit $x_1,...,x_k$. It's fairly easy that the whole orbit can't lie on one geodesic (otherwise it would contradict uniquely geodesic). Take the midpoints of the geodesic segments $[x_i,x_{i+1}]$ (mod $k$) and call them $x_i^{(1)}$. Do this inductively. Again by unique geodesic we have $d_j:=d(x_1^j,x_2^{j})>d_{j+1}$. From that follows easily by triangle inequality that $d(x_1^j, x_1^{j+1})$ is strictly decreasing. If this sequence converges to $0$, the $x_1^j$'s converge to a point which is $\phi$ invariant. But I have no idea why it should.
$X$ is contractible. If $\phi$ acts fixed point free, then $X\to X/\phi$ is a covering (is that even true?) that induces a metric on the quotient, which is a $K(\mathbb{Z}_k,1)$. Is it a contradiction that a $K(G,1)$ of a group having torsion is metrizable (e.g. because it's too complicated as a CW complex?
Give up and assume $X$ is moreover a manifold and find a compact convex $\phi$ invariant set. Show that such a set is homeomorphic to a closed ball (is that always true?) and use Brouwer.
