Consider a subgroup $H$ of the group $G = S_n \times \mathbb{Z}_2$ given by the direct product of some permutation group $S_n$ and the group $\mathbb{Z}_2$. Denote a general element of $G$ by $\pm \sigma$, where $\sigma \in S_n$.
Given a unitary irreducible matrix representation $\Gamma^\lambda_{\mu \nu}(\pm \sigma)$ of $H$ of symmetry $\lambda$, we can generate a "dual" representation $\Gamma^{-\lambda}_{\mu \nu}(\pm \sigma)$ by the prescription $$ \Gamma^{-\lambda}_{\mu \nu}(\pm \sigma) = \pm \Gamma^{\lambda}_{\mu \nu}(\pm \sigma) $$ This generates a new representation so long as their is some $-\sigma \in H$ for the character $\chi_\lambda(-\sigma)$ is non-zero. Moreover the representation is reducible since $$ \frac{1}{|H|}\sum_{\pm \sigma \in H} \chi^*_{-\lambda}(\pm \sigma) \chi_{-\lambda}(\pm \sigma) = \frac{1}{|H|}\sum_{\pm \sigma \in H} \left(\pm 1\right)^2 \chi^*_{\lambda}(\pm \sigma) \chi_{\lambda}(\pm \sigma) = 1 $$ I am curious about the multiplicities of $\lambda$ and $-\lambda$ in the reduction of an irreducible representation $\Gamma^{\lambda'}_{\mu'\nu'}$ of $G$. Since $G$ is a product group, its irreps are labelled by pairs of symmetries of its factors. Since the group $\mathbb{Z}_2$ has just two symmetries $\pm 1$, we can label the symmetries of $G$ by the symbols $\alpha \lambda'$, where $\alpha = \pm 1$ and $\lambda'$ is an irrep of $S_n$. We can likewise label the symmetries of $H$ by $\alpha \lambda$, keeping in mind that for "self-dual" representations we have $+\lambda \cong -\lambda$.
To determine the multiplicities we want to look at the inner product $\chi^\dagger_{\alpha \lambda}\chi_{\alpha' \lambda'} \equiv n_{\alpha \lambda, \alpha' \lambda'}$. First however, we note that, given a single "odd" element $-\sigma_o \in H$, we can write the group $H$ as the union of the disjoint sets $H_+$ and $-\sigma_o H_+ = H_-$, where $H_+ \subset H$ is the subgroup of "even" elements (i.e. pure permutations). Define in an analogous way the subgroup $G_+ \cong S_n$ and subset $G_-$. Looking now at the multiplicity formula, we have $$ \frac{1}{|H|}\sum_{\pm \sigma \in H} \chi_{\alpha \lambda}^*(\pm \sigma)\chi_{\alpha' \lambda'}(\pm \sigma) = \frac{1}{2|H_+|}\sum_{\sigma \in H_+} \left( \chi_{\alpha \lambda}^*(\sigma)\chi_{\alpha' \lambda'}(\sigma) + \chi_{\alpha \lambda}^*(- \sigma_o \sigma)\chi_{\alpha' \lambda'}(-\sigma_o \sigma) \right) $$ From here we note that $\chi_{\alpha' \lambda'}(-\sigma) = \alpha' \chi_{\lambda'}(\sigma)$ for all $-\sigma \in G_-$, so that $$ n_{\alpha \lambda, \alpha' \lambda'} = \frac{1}{2} \left( n_+ + \left(\alpha'\alpha\right) n_- \right) $$ where $$ n_\pm = \frac{1}{|H_\pm|}\sum_{\pm \sigma \in H_\pm} \chi^*_\lambda(\pm \sigma) \chi_{\lambda'}(\pm \sigma) $$ $n_+$ is incidentally the multiplicity of the irreps $\alpha \lambda$ of $H_+$ in the reduction of an irrep $\alpha' \lambda'$ of $G^+$ (it is independent of $\alpha$ and $\alpha'$).
That the multiplicity is unchanged under a simultaneous negation $\alpha, \alpha' \to -\alpha, -\alpha'$ I was able to determine independently. I want to know however what can be said under a single inversion $\alpha \to -\alpha$ or $\alpha' \to -\alpha'$. For the triply degenerate representations of $S_4$ and the subgroup $H \subset S_4 \times \mathbb{Z}_2$ comprised of the $\pm \sigma_\pm$ where $\sigma_+$ and $\sigma_-$ are respectively the even and odd permutations, we have that $n_- = \pm n_+$, the sign depending on how one selects which the assignment of signs to the pair of dual representations of $H$. Worth noting here is that in this particular case $H \cong S_4 \cong G_+$.
Does this relationship $n_- = \pm n_+$ hold in general? That is, if a non-self-dual irrep of a subgroup $H \subset G = S_n \times \mathbb{Z}_2$ appears in the reduction of an irrep $\alpha \lambda'$ of $G$, is it necessarily excluded from the reduction of the "dual" irrep $-\alpha \lambda'$?
A couple additional observations. In the case where $-e \in H$, then the $\alpha$ label in $\alpha \lambda$ can be used to classify the $\mathbb{Z}_2$ symmetry of the $H$ irrep and this symmetry is preserved on reduction so $\alpha \lambda$ can not appear in the reduction of $\alpha' \lambda'$ if $\alpha \neq \alpha'$. Another trivial example is when $H_+ = H$, in which case their are no non-self-dual irreps. So it suffices to consider the case where $H_+ \neq H$ and $-e \notin H$. Still this existence of a single "odd" element $-\sigma$ suffices to permit a set of representatives $g_i$, $i=1,\ldots,[G:H]$ of the cosets $G/H$ comprised of only "even" elements, i.e. $g_i \in G_+$ for all $i=1,\ldots,[G:H]$. This may be relevant.
