From topological actions on $\mathbb{R}^3$ to isometric actions 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
 A: My earlier attempt at an answer was a bit of a mess -- let me have another go.
The hypotheses of the question introduce several technical difficulties, but I'm unsure which are crucial and which can be relaxed. Certainly, if we're willing to relax them slightly then we can get a positive answer, so I'll give an answer under certain hypotheses that seem reasonable to me.
The usual discreteness hypothesis in this context is not an absence of accumulation points, but proper discontinuity, and it seems to me that this is the natural way to generalise Pardon's theorem.  With the hypothesis of absence of accumulation points (which @YCor rightly points out is strictly weaker) it's not clear to me what happens even for smooth actions. Apologies if this strictly weaker properness hypothesis is the point of the question (but I don't see a connection with Cannon's conjecture).
So let's suppose that $\Gamma$ is a hyperbolic group acting properly discontinuously and cocompactly by homeomorphisms on $\mathbb{R}^3$. To keep things simple, let's also assume that $\Gamma$ has a (wlog normal) torsion-free subgroup $\Gamma_0$ of finite index.
Since the action is properly discontinuous and $\Gamma_0$ is torsion-free, the action of $\Gamma_0$ is free and so the quotient $M_0=\Gamma_0\backslash\mathbb{R}^3$ is a closed topological 3-manifold.
By Moise's theorem $M_0$ has a smooth structure, and now $M_0$ is an aspherical 3-manifold whose fundamental group has no $\mathbb{Z}^2$ subgroups, so $M_0$ admits a hyperbolic metric by the geometrisation theorem. This metric pulls back to realise the action of $\Gamma_0$ as an action by isometries on $\mathbb{H}^3$.
Pardon's theorem shows that the action of the finite deck group $\Gamma\backslash\Gamma_0$ on $M_0$ can be approximated by smooth actions, and a theorem of Gabai implies that this action is isotopic to an action by isometries. As a result, the action of the whole group $\Gamma$ on $\mathbb{H}^3$ is also by isometries, as desired.
A: I believe (though have not checked carefully) that the argument in my paper proves:

If $\Gamma$ (discrete) acts continuously and properly discontinuously on a smooth three-manifold $M$, then that action can be uniformly approximated by a smooth action.

The point is simply that each step in the argument is local on the quotient space $M/\Gamma$ (which is a reasonable topological space given proper discontinuity).
Here is a (sketched) better argument, which proves the indented statement above as a consequence of my paper.  Fix $x\in M$, and consider the stabilizer $\Gamma_x\leq\Gamma$, which is finite.  Choose coset representatives $g_i\in\Gamma/\Gamma_x$, so $\Gamma x=\{g_ix\}_i$.  Fix a $\Gamma_x$-invariant open neighborhood $U$ of $x$ whose translates $g_iU$ are all disjoint (should exist by proper discontinuity).  Now smooth the action of $\Gamma_x$ on $U$ using my paper, and smooth the homeomorphisms $g_i:U\to g_iU$ using Bing--Moise.  This determines a smoothing of the action of $\Gamma$ on $\Gamma U\subseteq M$.  By making the approximations sufficiently $C^0$-close, we ensure that this smoothed action of $\Gamma$ on $\Gamma U\subseteq M$ splices together with the original action of $\Gamma$ on $M\setminus\Gamma U$ to define a new action of $\Gamma$ on $M$, which is now smooth over $\Gamma U$.  Now iterate a (locally) finite number of times to cover all of $M$.
