**Question**: Assume a smooth manifold $M$ does not admit any effective smooth group actions of finite groups $G \neq 1$, does it follow that $M$ also admits no *continuous* effective group actions of finite groups $G \neq 1$?

Manifolds which do not admit actions of finite groups are interesting because if $M$ is compact and we choose a Riemannian metric $g$ on $M$, then $\text{Isom}(M,g)$ is a compact Lie group by Steenrod-Myers and must be trivial, because otherwise it contains a non-trivial finite group acting on $M$.

For this nice property, it is enough to consider only smooth actions.

However, there are some interesting articles which construct manifolds which do not admit any *continuous* actions of finite groups.
And so I'm wondering if its enough to eliminate smooth actions if the manifold is smooth to get a manifold of this type.

If $\mathrm{Diff}^\infty(M)$ is torsion-free, does it follow that $\mathrm{Homeo}(M)$ is torsion-free?$\endgroup$ – YCor Jul 22 '19 at 14:32