 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, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.