= Solution
Set $n=6$ in part (b). The quotient $W=u^\perp/\langle u\rangle$ has dimension four and a <nondegenerate> alternating form. The coordinate action therefore defines
$$
\rho:S_6\longrightarrow Sp(W)\cong Sp(4,2).
$$
Prove it is faithful before comparing orders. If a <permutation> lies in the <kernel of a group homomorphism>, it fixes the quotient class of each two-element subset $A$. That class has exactly the two representatives $A$ and $A+\Omega=\Omega\setminus A$. Their cardinalities are two and four. A coordinate <permutation> preserves cardinality, so it must fix $A$ itself, not exchange it with its complement. Thus it fixes every two-element subset. Intersecting the fixed sets $\{i,j\}$ and $\{i,k\}$ shows it fixes each singleton $\{i\}$, so it is the identity. Hence $\rho$ is injective.
By part (a),
$$
|Sp(4,2)|=2^4(2^2-1)(2^4-1)=16\cdot3\cdot15=720=6!.
$$
An injection between <finite groups> of equal order is surjective. Choosing a <symplectic basis> of $W$ identifies the form with the standard one, so
$$
\boxed{Sp(4,2)\cong S_6.}
$$
This realizes the exceptional isomorphism through a concrete binary <permutation> module. Equality of orders alone would not establish it; the faithful form-preserving action supplies the required <group homomorphism>.
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