= Solution
Write the <group action> as $(\omega,g)\mapsto\omega g$. A right <group action> satisfies $\omega1=\omega$ and $(\omega g)h=\omega(gh)$. Its <orbit> and <stabilizer> are
$$
\operatorname{Orb}(\omega)=\{\omega g:g\in G\},\qquad H=\operatorname{Stab}(\omega)=\{g\in G:\omega g=\omega\}.
$$
The identity belongs to $H$. If $g,h\in H$, then $\omega gh=(\omega g)h=\omega$, so $gh\in H$. Also $\omega g=\omega$ implies $\omega=(\omega g)g^{-1}=\omega g^{-1}$, so $g^{-1}\in H$. Thus $H$ is a <subgroup>.
For the <orbit-stabilizer theorem>, the map $G\to\operatorname{Orb}(\omega)$, $g\mapsto\omega g$, is onto, and
$$
\omega g=\omega h\iff gh^{-1}\in H\iff g\in Hh.
$$
Its fibers are precisely the right <cosets> $Hh$, each containing $|H|$ elements. Counting these disjoint fibers gives
$$
\boxed{|\operatorname{Orb}(\omega)|=|G|/|H|}.
$$
For the <symmetric group> $S_3$, the six <subgroups> are
$$
\{1\},\quad \langle(12)\rangle,\quad\langle(13)\rangle,\quad\langle(23)\rangle,\quad A_3=\{1,(123),(132)\},\quad S_3.
$$
The <group action> $H\mapsto g^{-1}Hg$ has \b[four orbits]: each of $\{1\}$, $A_3$, and $S_3$ is fixed, while the three order-two <subgroups> form one <orbit>. Here an <orbit> is a collection of <subgroups>, rather than a collection of individual permutations.
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