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 $\phi:\mathbf{R}^n\to \mathbf{R}^n$ hashave a fixed point?
What I know about it:
If for every divisor $d | m$, the fixed point-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-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-order self-homeomorphisms.
Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that a certain power of it has its fixed point-point set very complicated.