It is known that if a finite group $G$ admits a faithful *topological* action on the 3-sphere $S^3$, then $G$ admits a faithful action on $S^3$ by *isometries*. (Pardon proved that a topological action implies a smooth action, and Dinkelbach & Leeb proved that a smooth action implies an isometric one.) I wonder if this extends to infinite groups acting on $R^3$:

**Question:** Let $G$ be a finitely generated group that admits a faithful, co-compact, topological action on $R^3$, such that no orbit has an accumulation point. Must $G$ admit an action by isometries on one of Thurston’s geometries, preserving the above properties (i.e. faithful, co-compact, accumulation-free)?

**Update:** The comments below suggest that the answer is negative in this generality (an "official" answer with references and explanation would be welcome). **What if G is assumed to be Gromov-hyperbolic?** I'm most interested in the 1-ended case, anticipating an isometric action on $\mathbb{H}^3$. (1-endedness excludes $\mathbb{S}^2 \times \mathbb{R}$.) This is partly motivated by Cannon's conjecture.

By topological action I mean an action by homeomorphisms.

**Update:** instead of just assuming that no orbit has an accumulation point, I'm happy with stronger discreteness conditions such as proper discontinuity