Let $M$ is a connected smooth $n$-manifold, and suppose that $f$ is a self-homeomorphism of $M$ that has finite order, i.e. $f^k = \text{id}$ for some $k\geq 1$. Suppose moreover that $f$ fixes a non-empty open set point-wise. Does it follow that $f = \text{id}$? In other words, is $\text{Homeo}(M,D^n)$ torsion-free?

If *homeomorphism* is replaced by *diffeomorphism*, this is easily seen to be true: create a Riemannian metric such that $f$ is an isometry (by averaging), then use the exponential map at one of the point whose tangent space is fixed by $f$ to deduce that $f = \text{id}$.

It is also true as stated for $M = S^n$; this can be proved using Smith theory.

But what about the general case?