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