= Solution
A <tabloid> is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\ldots,n$: order inside each row is forgotten, but the rows remain distinguished. The <permutation module> $M^\lambda$ has these tabloids as an $F$-<basis>, with $\Sigma_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and define
$$
\kappa_t=\sum_{\sigma\in C_t}\operatorname{sgn}(\sigma)\sigma,\qquad
e_t=\kappa_t\{t\}.
$$
The <Specht module> $S^\lambda$ is the <submodule> spanned by all polytabloids $e_t$.
If a <module> homomorphism $\phi:M^\lambda\to M^\mu$ is nonzero on $S^\lambda$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne\phi(e_t)=\kappa_t\phi(\{t\})$, so $\kappa_tM^\mu\ne0$. The allowed criterion yields $\lambda\ge\mu$.
In <characteristic zero>, <Maschke's theorem> gives an equivariant projection $M^\lambda\to S^\lambda$. If $S^\lambda\cong S^\mu$, compose that projection with the isomorphism and the inclusion into $M^\mu$. The preceding argument gives $\lambda\ge\mu$. Reversing the roles gives $\mu\ge\lambda$, so antisymmetry of dominance gives
$$
\boxed{\lambda=\mu.}
$$
For a map defined only on a <Specht module> in arbitrary characteristic, \b[dominance need not hold]. Take $F=\mathbb F_2$, $n=2$, $\lambda=(1,1)$ and $\mu=(2)$. The first <Specht module> is the one-dimensional span of
$$
e=\{1\mid2\}-\{2\mid1\}=\{1\mid2\}+\{2\mid1\}.
$$
The transposition fixes $e$, so this is the trivial <module>. The one-row <permutation module> $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single <tabloid> is a nonzero equivariant map, yet $(1,1)\not\ge(2)$. In this characteristic such a map need not extend to the surrounding <permutation module>; that is exactly where the earlier proof ceases to apply.
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