You can conclude that $|K|$ is a wedge sum of an odd number of $k+1$-dimensional spheres.

Proof: The case when $|K|$ is zero dimensional is obvious, so assume the dimension $k+1$ is greater than zero. By the Lefschetz fixed points theorem, if the action is fixed points free, then the trace of the action of the generator of ${\mathbb Z}/2$ on $H_{k+1}(K;{\mathbb Q})$ must be $(-1)^{k}$. This implies that the rank of $H_{k+1}(K;{\mathbb Q})$ is odd, since any representation of ${\mathbb Z}/2$ is a sum of copies of the trivial representation and the sign representation, so an even-dimensional representation will have an even trace.

On the other hand, it is easy to construct a free action of ${\mathbb Z}/2$ on a wedge sum of an odd number of spheres of any given dimension $d$. Imagine lining up the spheres in a row, letting them touch each other. Embed this union of spheres in ${\mathbb R}^{d+1}$ so that the center of the middle sphere lies at the origin. The antipodal action on ${\mathbb R}^{d+1}$ restricts to a free action on the wedge of spheres.

**Added later:** I wonder if this action is unique up to equivariant homotopy. My hunch is that it is, but I am not sure. One could begin by proving that the integral homology of $|K|$ is uniquely determined as an integral representation of ${\mathbb Z}/2$.