Happy easter to everyone!
I was wondering which kind of "local pictures" can arise when one quotients a manifold modulo the action of a finite group. I will call a $G$-manifold a space $X$ that can be obtained as the quotient $\tilde{X}/G$ for some space $\tilde{X}$.
Since this is a big question, I will try to cut the problem into smaller funnier pieces which can actually be answered.
Let me do an observation. If $(M, \partial M) $ is a manifold with boundary, then it is a $\mathbb{Z}_2$ - space. Indeed, if you glue two copies of $M$ along the boundary you will get a manifold: in a neighborhood of a point on the boundary, we have two copies of $\mathbb{R}^{n-1}\times \mathbb{R}_{\ge 0} $ glued along $\mathbb{R}^{n-1}\times \mathbb{R}_{= 0} $ , which is $\mathbb{R}^n$.
Question. Suppose $X$ is a smooth $\mathbb{Z}_2$-space, that is it can be written as $\tilde{X}/\mathbb{Z}_2$ with $\mathbb{Z}_2$ acting smoothly on the (smooth) manifold $\tilde{X}$.
- Are the fixed points $\tilde{X}^G$ a submanifold of $\tilde{X}$?
- Suppose $\tilde{X}^G$ is a submanifold of codimension 1. Is $X$ a manifold with boundary $\simeq \tilde{X}^G$?
First, let me make a comment about terminology. It is standard to use the term $\boldsymbol G$-space for a topological space equipped with a continuous action of a topological group $G$, and the term $\boldsymbol G$-manifold for a $G$-space that is also a manifold. Your space $\widetilde X$ on which $G$ acts is an example of a $G$-space, but the quotient space $\widetilde X/G$ is not (unless you give it the trivial $G$-action, which is not what you mean here); and since it's typically not a manifold at all, it's definitely not a $G$-manifold. You are of course free to introduce your own terminology for such a thing, but it's an exceedingly bad idea to use a term that has a well-accepted standard meaning and define it to mean something else. So you should think of another term for such a quotient space.
To address your specific questions about $\mathbb Z_2$ actions:
Yes. This MSE answer gives a proof.
Yes. If you follow the proof that I linked to above (or probably also the Klingenberg proof linked there), you'll see that in a neighborhood of each fixed point, the action by the nontrivial element of $\mathbb Z_2$ has a local coordinate represention of the form $(x^1,\dots, x^{n-1}, x^n) \mapsto (x^1,\dots, x^{n-1},-x^n)$. From there, it's easy to construct boundary coordinates for the quotient space. On the other hand, the fact that it's a diffeomorphism means that it maps a small coordinate neighborhood of each non-fixed point diffeomorphically onto a coordinate neighborhood of its image, and you can use this to construct an interior chart for the quotient in the same way one does for a free action.