 Solution

For $n\ge 3$, take $C_1$ to be the class of $n$-cycles, $C_2$ the class of transpositions, and $C_3$ the class of an $(n-1)$-cycle with one fixed point. With rightmost permutations applied first, a witness isMultiplying an $n$-cycle by a transposition splits it at the two exchanged letters. The resulting cycles have the lengths of the two arcs between those letters in the cyclic ordering. Thus $\alpha \beta$ has type $(n-1,1)$ exactly when the exchanged letters are cyclically adjacent.

Any product-one tuple can first be conjugated so that its first entry is $\alpha$. Its second entry is then one of the $n$ adjacent transpositions. The centralizer $⟨\alpha ⟩$ acts transitively on these choices; the third entry is forced to be the inverse product. Therefore every product-one tuple lies in a single simultaneous-conjugation orbit.

The conjugates ${\alpha }^j\beta {\alpha }^{-j}$ exchange all adjacent letters around the cycle. Their connected-edge graph generates $S_n$, so the witness, and every tuple in that orbit, generates $S_n$. HenceFor $n=3$ the last two classes coincide, which causes no difficulty: the same three adjacent swaps form one orbit and generate $S_3$.