Suppose $K\trianglelefteq N\trianglelefteq G$ are finite groups (where $K$ isn't necessarily normal in $G$), and suppose $\chi\in\mathrm{Irr}(N)$ is induced from $\mu\in\mathrm{Irr}(K)$. Let $V=\langle n\in N:\chi(n)\neq 0\rangle\trianglelefteq N$. Then it seems to me that the stabilizer of $\chi$ in $G$ can be computed as $$\mathrm{Stab}_G(\chi)=N\cdot \mathrm{Stab}_{N_G(V)}(\mu_V),$$ where $N_G(V)$ is the normalizer of $V$ in $G$ and $\mu_V$ is the restriction of $\mu$ to $V$. Is this correct?
Is there a better way to find the stabilizer in $G$ than finding $V$ and its normalizer, etc? I know that by Frobenius reciprocity I need to find the $g\in G$ for which $\langle \chi^g,\chi\rangle_G=\langle \chi^g_K,\mu\rangle_K\neq 0$, but I'm not sure how to decompose the restriction $\chi^g_K$ (maybe Mackey's restriction formula can be adjusted?). Thanks for help.