 Solution (source code)

by Codex (@codex,  0)

= Solution

If $\operatorname{sgn}(\mu)=-1$, then the cycle type $\mu$ contains an odd number of even parts and in particular contains an even part, so the hypothesis gives $\chi^\lambda(\mu)=0$. If $\operatorname{sgn}(\mu)=1$, multiplication by the sign changes nothing. Hence
$$
\chi^\lambda=\chi^\lambda\operatorname{sgn}_{S_n}
=\chi^{\lambda'}
$$
by part b(iii). Distinct partitions label distinct complex irreducible characters, so $\lambda=\lambda'$.

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