= Solution
The <Specht module> construction gives a nonzero inclusion $V^\lambda\hookrightarrow M^\lambda$ over the <complex numbers>: a <polytabloid> has coefficient one at its original <tabloid>, because the row and column stabilizers have trivial intersection, and its span is the corresponding <irreducible representation>. This establishes $a=M(\lambda,\lambda)\geq1$ without using the multiplicity-one conclusion. Similarly $a'=M(\lambda',\lambda')\geq1$.
By the corrected <tensor identity for induced representations> and the conjugate-partition sign twist, the multiplicity of $V^\lambda$ in $\widetilde M^{\lambda'}$ is $a'$. The <character inner product> already computed is the sum of products of multiplicities of common <irreducible representations>. Thus
$$
1=\langle\chi_{M^\lambda},\chi_{\widetilde M^{\lambda'}}\rangle\geq aa'\geq1.
$$
Both are positive <integers>, so $a=a'=1$. In particular
$$
\boxed{M(\lambda,\lambda)=1.}
$$
This deduction avoids assuming the full <Young's rule>, which would already contain the answer. Alternatively, the permitted one-dimensional <R-module homomorphism> space gives the same character inner product immediately.
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