**Question**: Let 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 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][1] 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 for this property.

  [1]: https://arxiv.org/pdf/math/0606714v1.pdf