= Solution
Expand the <Column antisymmetrizer of a Young tableau>:
$$
b_v\{w\}=\sum_{g\in C(v)}\operatorname{sgn}(g)\{gw\}.
$$
If two entries in one column of $v$ lie in the same row of $w$, their transposition belongs to both $C(v)$ and the row stabilizer of $w$, so the terms cancel in pairs. The assumption $b_v\{w\}\ne0$ therefore says that every row of $w$ meets every column of $v$ in at most one entry.
The first row of $w$ has $\lambda_1$ entries, while $v$ has exactly $\lambda_1$ nonempty columns. It must consequently contain exactly one entry from each column of $v$. Permuting within each column puts these entries in the first row positions of $v$. Delete the matched first rows and repeat the argument on the remaining <Young diagram>. The product of the resulting column permutations is an element $h\in C(v)$ for which the row sets of $hv$ are those of $w$. Thus
$$
\boxed{h\{v\}=\{w\}}.
$$
This is the <nonzero column antisymmetrizer criterion>.
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