 Solution

Let $s_i=(ii+1)$. Since $\mu -\mathrm{id}=s_i(\lambda -\mathrm{id})$, reindex the alternating sum defining ${\psi }^{\mu }$ by $\pi \mapsto s_i\pi$. This preserves all permutation-character terms and reverses every sign of a permutation, because $\mathrm{sgn}(s_i\pi )=-\mathrm{sgn}(\pi )$. Hence ${\psi }^{\mu }=-{\psi }^{\lambda }$.