Are fixed-point-free isometries closed under composition?

Question

Question: Are fixed-point-free isometries closed under composition?

Clarification: I want to exclude the trivial case that the composition of a fixed-point-free isometry with its inverse is the identity. In other words, if the composition does not equal the identity, is it still fixed-point-free? Or do fixed-point-free isometries form a group for an arbitrary metric space?

Context: This might reduce to a group-theoretic question, since the isometries of a metric space are just a subgroup of the "permutation" group of the space (group of self-maps which are bijections). I am not sure how to tackle a problem with infinite permutation groups however.

Note that this result is true for Euclidean space -- the composition of two translations is again either a translation or the identity, and the translations are exactly the fixed-point-free isometries.

However, I think it is false for general metric spaces, but cannot come up with a counterexample.

Answer 1

It is false for a general metric space. Specifically, let $X$ be a space with $5$ points and the discrete metric, and look at an isometry that swaps two points and cycles the other three. It's composition with itself is not the identity, but it has fixed points.

Answer 2

The result is false even for Euclidean spaces; identifying the Euclidean plane with the complex plane, we have fixed point free isometries given by $$z\ \longmapsto\ \overline{z}+1\qquad\text{ and }\qquad z\ \longmapsto\ z-(1+i),$$ but their composition $z\ \longmapsto \overline{z}-i$ has the fixed point $z=-\frac{i}{2}$.