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Because $\lambda$ is self-conjugate, every off-diagonal hook of even length is paired with its transpose, while every diagonal hook has odd length. Suppose even hooks exist and let $L$ be their maximum length. Apply the Murnaghan–Nakayama rule to cycle types beginning with $L$ and complete the remaining cycle type with the principal hooks of the residual diagram. The principal-hook character value of a symmetric group makes each surviving residual character equal to $1$ or $-1$.

The assumed vanishing forces cancellation among the removable $L$-hooks. The standard maximal-hook comparison shows that the only possible cancellation is one transposed pair: the hooks must be $H_{1,j}$ and $H_{j,1}$ for a single $j>1$. Any further hook of length $L$, or a maximal hook with both indices greater than one, can be isolated by the residual principal-hook cycle type and would give a nonzero value. Thus either there are no even hooks or the maximum even length occurs exactly at that pair.

Choose $g$ whose disjoint permutation cycles have lengths equal to the principal hook lengths of $\lambda$. Those lengths are distinct and sum to $n$, so this is a permutation in $S_n$. In the iterated Murnaghan–Nakayama rule, there is a unique complete sequence that removes the corresponding principal rim hooks. Its contribution is one sign, and hence the Principal-hook character value of a symmetric group gives ${\chi }^{\lambda }(g)\in \{1,-1\}$.