Existence of Fundamental Domain on $\mathbb{R^{2}}$

Question

Consider a group G acting on the plane. Does it always have a fundamental domain?

I tried to break it in cases and here I present my results;

If $G$ is an isometry group acting discontinuously then I know its generated by two elements and it always have a fundamental domain,in fact I know how to calculate the quotient space.

If $G$ is an isometry group action continuosly then I know there is some $G$ such that I can calculate the fundamental domain.

If $G$ is not a group of isometries I have no idea of how to approach it and I appreciate any hints or theorems that helps me to prove/disprove it.

Edit: By fundamental domain I mean a region $A\subseteq\mathbb{R^{2}}$ such that the interior of $A$ contains exactly one point from each orbit and given any point $p\in\mathbb{R^{2}}$ , $p=gx$ for some $x$ in the closure of $A$ and $g\in G$.

Answer 1

It a general case an answer is clearly negative. For instance, when $G$ is a group of all bijections of the plane, only one-point sets contain one point from each orbit. You need some restrictions on the action of $G$ (I guess, of continuity type) in order to obtain a positive answer. Also there should be no dense orbits. I guess that a typical orbit should be countable and discrete.

PS.

the interior of $A$ contains exactly one point from each orbit and given any point $p\in\mathbb{R^{2}}$ , $p=gx$ for some $x$ in the closure of $A$ and $g\in G$.

If the interior of $A$ contains exactly one point from each orbit then already given any point $p\in\mathbb{R^{2}}$ , $p=gx$ for some $x$ in the interior of $A$ and $g\in G$.