Take a contractible manifold $C$, multiply it by $\mathbb R^n$, and let the finite group act trivially on $C$, and linearly on $\mathbb R^n$ such that $0$ is the unique fixed point. Then $C\times 0$ is the fixed point set. If $C$ is not diffeomorphic to $\mathbb R^n$, the action is not linear, but for sufficiently large $n$ the product $C\times\mathbb R^n$ will be diffeomorphic to a Euclidean space, by Stallings's characterization of Euclidean space as the contactible space that is simply-connected at infinity. In fact, Craig Guilbault proved that $n=1$ suffices (except possibly when $\dim(C)=3$, which I do not quite understand at the moment). See here for Craig's paper.

Take a contractible manifold $C$, multiply it by $\mathbb R^n$, and let the finite group act trivially on $C$, and linearly on $\mathbb R^n$ such that $0$ is the unique fixed point. Then $C\times 0$ is the fixed point set. If $C$ is not diffeomorphic to $\mathbb R^n$, the action is not linear, but for sufficiently large $n$ the product $C\times\mathbb R^n$ will be diffeomorphic to a Euclidean space, by Stallings's characterization of Euclidean space as the contactible space that is simply-connected at infinity. In fact, Craig Guilbault proved that $n=1$ suffices (except possibly when $\dim(C)=3$, which I do not quite understand at the moment). See here for Craig's paper.