= Solution
Use the six-point matching geometry. A <duad> is an edge of the <complete graph> on six points, a <syntheme> is a <perfect matching> of its three pairs, and a <pentad> is a one-factorization: five pairwise edge-disjoint <synthemes> covering all fifteen <duads>.
First count the <pentads> without assuming their number. There are fifteen <synthemes>: the partner of the first point has five choices, the least remaining point then has three possible partners, and the last pair is forced. For a fixed <syntheme> $M$, eight others share no edge with it: <inclusion-exclusion> gives $15-3\cdot3+3\cdot1-1=8$. Two edge-disjoint <synthemes> $M,M'$ have union a six-cycle. The complement of that cycle is a triangular prism, consisting of two triangles with three corresponding cross edges. It has precisely four <perfect matchings>: the all-cross matching and three matchings using one cross edge plus one edge from each triangle. The all-cross matching leaves two odd triangles and therefore cannot be completed by two matchings. The other three matchings are pairwise edge-disjoint and partition the prism. Hence $M,M'$ extend to exactly one <pentad>.
Each <pentad> through $M$ uses four of its eight disjoint partners, and every partner gives exactly one completion. Thus $M$ lies in two <pentads>. Counting incidences gives $15\cdot2/5=6$ <pentads> in total. Distinct <pentads> cannot share two <synthemes> because two disjoint <synthemes> determine their unique completion. Every <syntheme> therefore identifies a distinct pair of <pentads>; there are fifteen such pairs, so this incidence map is a bijection.
Permuting the six original points permutes the six <pentads>, defining a <group homomorphism> $\Phi:S_6\to\operatorname{Sym}(\text{pentads})$. It is faithful. If an element fixes every <pentad>, it fixes every <syntheme> by the pair-incidence bijection. A <duad> is the unique common edge of two distinct <synthemes> containing it, so every <duad> is fixed as well. A <permutation> fixing all two-element subsets fixes every point, by intersecting $\{i,j\}$ and $\{i,k\}$ for distinct $i,j,k$. The <kernel of a group homomorphism> is therefore trivial. Both <groups> have order $6!$, so labeling the <pentads> produces an <group automorphism> of $S_6$.
It remains to prove this <group automorphism> is outer. A <transposition>, say $(1\ 2)$, fixes no <pentad>. If it fixed one, the unique <syntheme> in that <pentad> containing the <duad> $\{1,2\}$ would be fixed. Every other <syntheme> $M'$ in the <pentad> contains 1 and 2 in separate edges. Swapping them changes those two edges but leaves the third edge unchanged. Hence $(1\ 2)M'$ is a distinct <syntheme> sharing an edge with $M'$, impossible for two members of the same <pentad>. Thus $(1\ 2)$ acts without fixed points on the six <pentads> and becomes a product of three disjoint <transpositions>.
<Inner automorphisms> preserve <cycle type>, whereas $\Phi$ sends a single <transposition> to a triple <transposition>. Consequently
$$
\boxed{\Phi\text{ is an outer automorphism of }S_6.}
$$
This realizes the exceptional class-size coincidence from part (a). It is the full symmetric-group version of the <pentad construction of the exceptional alternating-group automorphism>, with faithfulness proved directly from incidence rather than borrowed from a simplicity theorem.
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