 Solution

If $\mathrm{sgn}(\mu )=-1$, 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 $\mathrm{sgn}(\mu )=1$, multiplication by the sign changes nothing. Henceby part b(iii). Distinct partitions label distinct complex irreducible characters, so $\lambda ={\lambda }^{\prime }$.