Conjugating a tableau exchanges row symmetrization with column antisymmetrization. The resulting module is the original Specht module twisted by the <sign representation>, giving the <conjugate Specht character>
$$
S^{\lambda'}\cong S^\lambda\otimes\operatorname{sgn}_{S_n}.
$$
Taking traces yields
$$
\chi^{\lambda'}=\chi^\lambda\operatorname{sgn}_{S_n}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 2 / b / iii / Solution

Conjugating a Young diagram twists its complex Specht module by the sign representation:
$$
S^{\lambda'}\cong S^\lambda\otimes\operatorname{sgn},
\qquad
\chi^{\lambda'}=\chi^\lambda\operatorname{sgn}.
$$


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