I want to know if every smooth (finite)group action on $\mathbb{R}^n$ is conjugate to some linear action.Thank you!
Yes in dimensions ≤ 2 (classical). Yes in dimension 3 via the Geometrization Conjecture (with much earlier work in special cases). No in higher dimensions, with the simplest examples perhaps being counterexamples to the "Smith conjecture"smooth involutions on R^4 (or S^4) with knotted 2dimensional fixed point set (Giffen, Gordon, Sumners). 


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 simplyconnected 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. 


For the algebrogeometric setting, that is polynomial automorphisms (and conjugacy in the group of polynomial automorphisms), I recall a talk and several surveys by Hanspeter Kraft where it was stated that it is true for $n=1,2$ and unknown for $n>2$. I am not sure if it gives reasonable intuition for smooth automorphisms though. 

