= Solution
Put $W=S_F^{(n-2,1,1)}$ and let $H\cong S_{n-1}$ be the <subgroup> fixing one. For $\tau=(r,s)\in H$, let $u=\min\{\tau(i),\tau(j)\}$ and $v=\max\{\tau(i),\tau(j)\}$. Permuting a tableau's entries and then ordering its first column gives
$$
\boxed{\tau e_{1ij}=\begin{cases}e_{1uv},&\tau(i)<\tau(j),\\-e_{1uv},&\tau(i)>\tau(j).\end{cases}}
$$
The underlying <basis> element is $e_{1ij}$ exactly when $\tau$ preserves the unordered pair $\{i,j\}$: either $\{r,s\}=\{i,j\}$, giving the minus sign, or the two pairs are disjoint, giving the plus sign.
If $p\neq2$, invariance of $x$ under $(i,j)$ forces its $e_{1ij}$ <coefficient> to satisfy $a_{ij}=-a_{ij}$, because no other <basis vector> is sent to that <basis vector>. Thus every <coefficient> vanishes. Consequently $W^H=0$, and in particular $W^{S_n}=0$, so there is no nonzero <homomorphism> from the trivial <module>. Any <homomorphism> $M_F^{(n-1,1)}\to W$ is determined by the <image> of its <tabloid> with singleton row $\{1\}$. That <image> is fixed by $H$ and is zero, making the whole <homomorphism> zero.
\b[The printed vanishing assertion is false in <characteristic two>.] Signs then coincide, so $H$ permutes the <basis> indexed by unordered pairs from $\{2,\ldots,n\}$. This action is <transitive>, and
$$
\boxed{W^H=Fv,\qquad v=\sum_{2\leq i<j\leq n}e_{1ij}\neq0\quad(p=2).}
$$
The vector is nonzero by standard-basis independence. It also gives a nonzero map from the <permutation module>: send the singleton-row <basis vector> indexed by $r$ to $g_rv$, where $g_r(1)=r$. This is well defined because two choices of $g_r$ differ by an element of $H$, and it is $S_n$-equivariant by construction. Thus the claimed vanishing of permutation-module <homomorphisms> fails for every $n\geq4$ in <characteristic two>.
The full-group fixed space can be determined explicitly. Denote a <hook of a Young diagram>[hook] <tabloid> by $[a,b]$, recording its second- and third-row entries. In <characteristic two>, $e_{1ij}$ is the sum of the six <tabloids> using ordered pairs from $\{1,i,j\}$. In $v$, a <tabloid> whose two recorded entries both differ from one appears once; a <tabloid> using one appears $n-2$ times. Hence for odd $n$, $v$ is the sum of all <hook of a Young diagram>[hook] <tabloids> and is fixed by $S_n$. For even $n$, it is the sum of precisely the <tabloids> avoiding one and is moved by $(1,2)$. Since every full-group fixed vector already lies in $Fv$, the corrected <dimensions> are
$$
\boxed{\begin{array}{c|cc}
&\dim W^H&\dim W^{S_n}\\\hline
p\neq2&0&0\\
p=2,\ n\text{ even}&1&0\\
p=2,\ n\text{ odd}&1&1
\end{array}}
$$
Thus even the trivial-module vanishing fails for odd $n$ in <characteristic two>; for example $n=5$ gives the explicit nonzero <invariant> $v$. These corrected fixed spaces allow the valid later conclusions to be proved in all characteristics.
The corrected parity dependence is described by <fixed vectors in a hook Specht module in characteristic two>.
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