Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ has a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite order self-homeomorphisms.

Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that certain power of it has fixed point set very complicated.