= 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> are
$$
\boxed{\operatorname{Orb}(\omega)=\{g\omega:g\in G\},\qquad
G_\omega=\{g\in G:g\omega=\omega\}.}
$$
The <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> $gG_\omega$ 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 $|\operatorname{Orb}(\omega)|=[G:G_\omega]=|G|/|G_\omega|$ for finite $G$.
For <conjugation> in $S_3$, the <group orbits> are exactly the conjugacy classes, determined by <cycle type>:
$$
\boxed{\{1\},\qquad\{(12),(13),(23)\},\qquad\{(123),(132)\}.}
$$
<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 $6,2,3$.
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