 Solution

A group action is a map $G\times \Omega \to \Omega$, $(g,\omega )\mapsto g\omega$, satisfying $1\omega =\omega$ and $(gh)\omega =g(h\omega )$. Its group orbit and stabilizer areThe stabilizer contains the identity. If $g,h$ fix $\omega$, then $gh$ fixes it; and $g\omega =\omega$ implies $g^{-1}\omega =\omega$ after applying $g^{-1}$. Thus it is a subgroup.

The map from left cosets $g{G}_{\omega }$ to group-orbit points $g\omega$ is well-defined, surjective and injective: $g\omega =h\omega$ is equivalent to $h^{-1}g\in G_{\omega }$. Therefore $|\mathrm{Orb}(\omega )|=[G:G_{\omega }]=|G|/|G_{\omega }|$ for finite $G$.

For conjugation in $S_3$, the group orbits are exactly the conjugacy classes, determined by cycle type:Conjugation sends $(i_1\cdots i_k)$ to $(g(i_1)\cdots g(i_k))$, so these displayed sets are invariant and each is one group orbit. Their sizes $1,3,2$ also agree with centralizer orders $6,2,3$.