= Solution
The row stabilizer of the transposed tableau is $R(t')=C(t)$. For $r\in R(t')$, the <polytabloid> satisfies
$$
r e(t)=\operatorname{sgn}(r)e(t),
\qquad
r e(u)=\operatorname{sgn}(r)e(u).
$$
The two signs cancel in the tensor product, so
$$
gr e(t)\otimes gr e(u)=g e(t)\otimes g e(u).
$$
Thus the proposed value depends only on the tabloid $\{gt'\}$ and $\theta$ is well-defined. Its definition immediately gives
$$
\theta(k\{gt'\})=k\theta(\{gt'\}),
$$
so it is an $FS_n$-homomorphism. Since any $e(w)$ is $g e(t)$ for some $g$, its images contain every generator $e(w)\otimes e(u)$ of $S^\lambda\otimes S^{(1^n)}$. Hence $\theta$ is surjective.
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