
  Suppose that $G$ is a finite group, acting via homeomorphisms on $B^n$, the closed $n$-dimensional ball.  Does $G$ have a fixed point?


A fixed point for $G$ is a point $p \in B^n$ where for all $g \in G$ we have $g\cdot p = p$.
Notice that the answer is "yes" if $G$ is cyclic, by the Brouwer fixed point theorem.  Notice that the answer is "not necessarily" if $G$ is infinite.  If it helps, in my application I have that the action is piecewise linear.  

First I thought this was obvious, then I googled around, then I read about Smith theory, and now I'm posting here. 

The answer is no.
A fixed point free action of the finite group $A_5$ on a $n$-cell was constructed by Floyd and Richardson in their paper An action of a finite group on an n-cell without stationary points, Bull. Amer. Math. Soc. Volume 65, Number 2 (1959), 73-76.
For some non-existence results, you can see the paper by Parris Finite groups without fixed-point-free actions on a disk, Michigan Math. J. Volume 20, Issue 4 (1974), 349-351.
