= Solution
The relevant cycle types are $(5,1)$, $(2,2,2)$ and $(3,2,1)$. Conjugate the first entry to $\alpha=(1\,2\,3\,4\,5)$, fixing six. An entry $\beta$ of type $(2,2,2)$ pairs six with one of the five other letters. Powers of $\alpha$ move that partner to one. With the pair $(1\,6)$ fixed, only three matchings remain:
$$
\begin{array}{c|c}
\beta&\text{cycle type of }\alpha\beta\\
(1\,6)(2\,3)(4\,5)&(4,1,1)\\
(1\,6)(2\,4)(3\,5)&(6)\\
(1\,6)(2\,5)(3\,4)&(3,2,1).
\end{array}
$$
Thus there is exactly one <centralizer> orbit of eligible second entries, consisting of five matchings before normalization. The third entry is forced to be $(\alpha\beta)^{-1}$, so there is at most one inner orbit of product-one tuples of these types.
For the actual displayed $g_2$, direct multiplication gives
$$
\alpha g_2=(1\,3)(4\,6\,5),\qquad
(\alpha g_2)^{-1}=(1\,3)(4\,5\,6),
$$
which belongs to the class of the displayed $g_3$. This is a product-one witness. The three displayed <permutations> themselves need not multiply to one; the problem concerns their conjugacy classes.
The group $\langle\alpha,g_2\rangle$ is transitive, because $\alpha$ moves the first five points transitively and $g_2$ connects six to them. Its stabilizer of six contains the five-cycle $\alpha$ and is transitive on the other points. Hence the group is <two-transitive> and therefore primitive. Moreover $(\alpha g_2)^3=(1\,3)$ is a <transposition>. The primitive <transposition> lemma proves that the group is $S_6$. Consequently all tuples in the sole inner orbit generate $S_6$, and
$$
\boxed{(C_{(5,1)},C_{(2,2,2)},C_{(3,2,1)})\text{ is rigid in }S_6.}
$$
As a count check, there are $6\cdot4!=144$ first entries and five eligible second entries for each, giving $720$ tuples. A generating tuple has simultaneous stabilizer equal to the trivial center of $S_6$, so its inner orbit also has size $720$.
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