= Solution
Write $a=\lambda-\mathrm{id}$. The two modified entries satisfy
$$
\mu_i-i=\lambda_{i+1}-(i+1)=a_{i+1},
\qquad
\mu_{i+1}-(i+1)=\lambda_i-i=a_i.
$$
Thus $\mu-\mathrm{id}$ is obtained from $\lambda-\mathrm{id}$ by the adjacent transposition $\tau=(i,i+1)$. In the alternating definition of $\psi$, reindexing $\pi$ by $\tau\pi$ preserves every induced permutation character and reverses every sign. Therefore the <straightening of a symmetric-group character indexed by a composition> gives
$$
\boxed{\psi^\mu=-\psi^\lambda}.
$$
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