Suppose $H$ is a finite group acting smoothly on a smooth connected manifold $M$. The action is trivially proper, as $H$ is discrete. If the action of $H$ were also known to be free, i.e. $h\cdot p\neq p$ for all $p\in M$ and non-identity $h\in H$, then it would follow that the quotient space $M/H$ by the action of $H$ would have a unique structure of a manifold making the quotient map a a smooth normal covering.
My question is, without knowing a priori that the action of $H$ is free, does the finiteness of $H$ ensure that the action is free? If not, can we say something maybe less strong but to the same effect, namely something admitting a smooth normal covering map?
My idea was to somehow define define a different action of $H$ on $M$ which was smooth and free, thereby obtaining a quotient map which, if I was clever enough, intertwined the $H$ actions, but I am not even sure this is possible. Any hints?