TITLE: Existence of equivariant triangulations
QUESTION [5 upvotes]: I'm looking for a compatibility result which links two types of structures that could be imposed on a topological space $X$:

Call $X$ triangulable if there exists a finite simplicial complex $K$ whose geometric realization $|K|$ admits a homeomorphism to $X$.
Call $X$ involutive if it admits a nontrivial $\mathbb{Z}/2$-action, in the sense that there is a continuous $\iota:X \to X$ satisfying $\iota \circ \iota = \text{id} \neq \iota$.


Here is the question: let's say a given space $X$ admits a triangulation $K$ and an involution $\iota$ in the senses described above. Is it true that we can find another triangulation $K'$, ideally a subdivision of $K$, with respect to which $\iota$ is a simplicial map?

There are more general versions of this question about which I'm curious, obtained by replacing $\mathbb{Z}/2$ by another finite group $G$ acting on a triangulable $X$. If the desired statement is true, I would love to be pointed to a reference; and if it is false, I wonder if there is a nice obstruction to the existence of $K'$.

REPLY [3 votes]: There are involutions $\sigma$ of the 3-sphere, whose fixed-point sets are wild 2-spheres: The fixed-point set cannot be a subcomplex of any triangulation, hence, $\sigma$ cannot be PL in any triangulation.
Bing, R. H., A homeomorphism between the 3-sphere and the sum of two solid horned spheres, Ann. Math. (2) 56, 354-362 (1952). ZBL0049.40401.
See also here for Calegari's take on Bing's proof.
Edit. There is even (unique in some sense) free involution of the 4-sphere which cannot preserve a triangulation (this is due to Ruberman). Thus, a bad fixed point set is not the only obstruction.