 Degree-ten partition action of S6

There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions of six letters into two unordered triples. A stabilizer is $S_3\wr S_2$, also the normalizer of the Sylow subgroup generated by one 3-cycle in each triple. Relative to one partition, all other partitions have intersection pattern $1+2$ up to interchange, giving one orbit of the stabilizer and hence a two-transitive action. A natural $S_5$ fixing one letter is transitive on these partitions, giving $S_6=S_5(S_3\wr S_2)$.