= Solution
The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two <double cosets>
$$
S_n=H\sqcup HtH,\qquad t=(n-1\ n).
$$
An $H$-bimodule-equivariant set map is determined by $u=\Phi(1)$ and $v=\Phi(t)$. Equivariance requires $u\in Z(H)$ and
$$
v\in C_H(S_{n-2}),
$$
because $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi(h)=hu$ and $\Phi(hth')=hvh'$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.
For $n\geq5$, the center of $H$ and this <centralizer> are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\ldots,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\geq3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(1\ 2)\}$. The choice
$$
\Phi(h)=h,\qquad \Phi(h(3\ 4)h')=h(1\ 2)h'
$$
is well defined and satisfies all the printed equivariance conditions, but sends $(3\ 4)$ to $(1\ 2)$ rather than to the identity. \b[Uniqueness holds for $n\geq5$; it is false at $n=4$ as printed.] Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.
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