= Solution
The <symmetric group> $S_3$ has <subgroups> of orders $1,2,3,6$. There is one trivial <subgroup>, three conjugate <subgroups> generated by transpositions, one order-three <subgroup> $A_3$, and the whole <group>. These are all <subgroups> by <Lagrange's theorem>. Thus again there is one conjugacy class for each index $6,3,2,1$.
The <coset action> classification in part (c) supplies exactly one transitive <orbit> type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, giving
$$
\boxed{8\text{ inequivalent actions}.}
$$
The equality of the two counts comes from their identical lists of transitive <orbit> types, rather than from the <groups> being isomorphic.
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