= Solution
For $n\geq3$, 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 is
$$
\alpha=(1\,2\,\cdots\,n),\qquad\beta=(1\,2),\qquad
\gamma=(\alpha\beta)^{-1}.
$$
Multiplying 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> $\langle\alpha\rangle$ 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 $S_n$. Hence
$$
\boxed{(C_{(n)},C_{(2,1^{n-2})},C_{(n-1,1)})\text{ is a rigid triple in }S_n.}
$$
For $n=3$ the last two classes coincide, which causes no difficulty: the same three adjacent swaps form one orbit and generate $S_3$.
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