= Solution
It suffices to handle one <single-box up-move>, and then use the chain already constructed. Suppose the affected row sizes are $a=\lambda_i$ and $b=\lambda_j$, $i<j$. Since $\lambda$ is a <partition of an integer>, $a\geq b\geq1$. Put $N=a+b$. Over the <complex numbers>, the <two-row Young permutation module decomposition> gives
$$
M^{(a,b)}\cong\bigoplus_{r=0}^b V^{(N-r,r)},\qquad
M^{(a+1,b-1)}\cong\bigoplus_{r=0}^{b-1}V^{(N-r,r)}.
$$
A zero second row is omitted. Therefore
$$
M^{(a,b)}\cong M^{(a+1,b-1)}\oplus V^{(a,b)}.
$$
Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant <Young subgroups> lie in $K$. Transitivity of <induced representations> expresses $M^\lambda$ and $M^\nu$ by inducing the preceding two-row modules, tensored with the <trivial representations> of the unaffected factors, from $K$ to $S_n$. Induction preserves this <direct sum>, so
$$
M^\lambda\cong M^\nu\oplus L_{\lambda,\nu}
$$
for an actual <group representation> $L_{\lambda,\nu}$, not merely a difference of characters. Iterate along the chain and take the <direct sum> of the induced complements. The resulting complement can be realized as an <invariant subspace> of $M^\lambda$, either through these isomorphisms or by <Maschke's theorem>. Thus
$$
\boxed{M^\lambda\cong M^\mu\oplus L_{\lambda,\mu}.}
$$
If $\lambda=\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.
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