= Solution
Let $0\ne f:S^\lambda\to M^\mu$ be a module homomorphism. Since $S^\lambda$ is simple, $f$ is injective. Fix a $\lambda$-tableau $t$; its nonzero <polytabloid> $e_t$ then has $f(e_t)\ne0$. The sign sum in the column group satisfies
$$
\kappa_t^2=|C_t|\kappa_t,
$$
because every coefficient in the square receives $|C_t|$ equal contributions. Thus $\kappa_te_t=|C_t|e_t$, and equivariance gives
$$
\kappa_tf(e_t)=f(\kappa_te_t)=|C_t|f(e_t)\ne0.
$$
The <column antisymmetrizer> therefore acts nontrivially on $M^\mu$, so some basis <tabloid> $[s]$ has $\kappa_t[s]\ne0$. The stated <dominance from a nonzero column antisymmetrizer> yields
$$
\boxed{\operatorname{Hom}_{\mathbb CS_n}(S^\lambda,M^\mu)\ne0\Longrightarrow\lambda\unrhd\mu.}
$$
For $\mu=\lambda$, the same calculation forces $f(e_t)=|C_t|^{-1}\kappa_tf(e_t)$ into the one-dimensional space $\mathbb Ce_t$. Write $f(e_t)=ce_t$. Since $e_t$ generates the <Specht module>, equivariance gives $f(ge_t)=cge_t$ for every $g$, so $f$ is $c$ times the inclusion $S^\lambda\hookrightarrow M^\lambda$. That inclusion is nonzero, hence
$$
\boxed{\dim\operatorname{Hom}_{\mathbb CS_n}(S^\lambda,M^\lambda)=1.}
$$
As a useful consequence, if $S^\lambda\cong S^\mu$, their inclusions into $M^\lambda,M^\mu$ give dominance both ways. Antisymmetry of the <dominance order on partitions> then gives $\lambda=\mu$, so different partitions produce nonisomorphic simple modules.
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