= Degree-ten partition action of S6
{title2=$S_6\curvearrowright\{\text{unordered }3+3\text{ partitions}\}$}
There are $\binom63/2=10$ 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 $S_6=S_5(S_3\wr S_2)$.
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