 Solution

The relevant cycle types are $(5,1)$, $(2,2,2)$ and $(3,2,1)$. Conjugate the first entry to $\alpha =(12345)$, 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 $(16)$ fixed, only three matchings remain: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 giveswhich 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 $⟨\alpha ,g_2⟩$ 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 $(\alpha g_2{)}^3=(13)$ is a transposition. The primitive transposition lemma proves that the group is $S_6$. Consequently all tuples in the sole inner orbit generate $S_6$, andAs a count check, there are $6\cdot 4!=144$ 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 $720$.