Let $G$ be a group of bijective transformations of a set $X$. For all $x\in X$, the orbit of $x$ is the set $$G(x)=\{gx:g\in G\},$$ where $gx\in X$ is the result of the action of $g$ on $x$.
A set $A\subseteq X$ is said invariant if $gA=A$ for all $g\in G$, where $gA=\{gx:a\in A\}$. Any orbit $G(x)$ is clearly invariant, and any invariant set is an arbitrary union of orbits. Additionally, the set $I$ of all invariant sets is a $\sigma$- algebra.
My question is the following. Consider the $\sigma$-algebra $O$ generated by all orbits $G(x)$. I found stated elsewhere (without proof) that $I=O$, but I can't seem to prove this. The inclusion $O\subseteq I$ is trivial, but I am not sure how to proceed with the reverse inclusion. Does anyone know if the equality $I=O$ is actually true, and if so how to prove it?
This is not true in general. Indeed, let $A$ be the collection of sets which are either a countable union of orbits or the complement of a countable union of orbits. It is not hard to see that $A$ is a $\sigma$-algebra. Since every orbit is an element of $A$, $O\subseteq A$. But if there are uncountably many orbits, then not every invariant set is in $A$ (split the orbits into two uncountable sets and take the union of one of them), so $O\neq I$.
(Conversely, if there are only countably many orbits, then every invariant set is a countable union of orbits so $O=I$. So $O=I$ iff there are only countably many orbits.)