$\varphi^G$=$\psi^G$ implies that subgroups are conjugate

Question

Let $G$ be a finite group and let $\varphi,\psi$ be linear characters of $H$ and $K$ respectively, where $H,K$ are subgroups of $G$.

I'm trying to prove that if the $G$-induced characters of $\varphi,\psi$ are equal (this is, $\varphi^G=\psi^G$) then $H$ and $K$ are conjugate in $G$. Yet I don't know if the result is true either. If it was true, it would conclude another proof that I'm attempting.

If $\varphi^G=\psi^G$, evaluating the function at $1$ we obtain that $|G:H|=|G:K|$ and hence $|H|=|K|$. This makes me think that the result may be true.

I have also proved that $\psi_{H^s\cap K}=\varphi^s_{H^s\cap K}$ for some $s\in G$. Mackey's decomposition formula states that if $S$ is a set of double coset representatives so that $G=\cup_{s\in S}HsK$ is a disjoint union then $$(\varphi^G)_K=\sum_{s\in S}(\varphi^s_{H^s\cap K})^K.$$

Now, by Frobenius reciprocity it holds $0\not=[\varphi^G,\psi^G]=[(\varphi^G)_K,\psi]=\sum_{s\in S}[(\varphi^s_{H^s\cap K})^K,\psi]=\sum_{s\in S}[\varphi^s_{H^s\cap K},\psi_{H^s\cap K}]$. Thus, there exists $s\in S$ such that $0\not=[\varphi^s_{H^s\cap K},\psi_{H^s\cap K}]$ and hence $\psi_{H^s\cap K}=\varphi^s_{H^s\cap K}$ (as both characters are linear and therefore irreducible).

However, I don't know neither this equality of characters may be relevant, as $H^s\cap K$ may be trivial.

Any hint or suggestion for how to proceed is appreciated.

Answer 1

There appears to be some literature around the study of such pairs of subgroups, either just named "subgroups inducing the same permutation representation" like in this article titled, well, Subgroups inducing the same permutation representation by Robert M. Guralnick, or "linked subgroups" in this paper titled Subgroups of finite soluble groups inducing the same permutation character by Norberto Gavioli, and more references are given in the introduction of Gavioli's paper.

Guralnick's theorem A seems to answer your claim by the negative by providing the existence of counterexamples under some conditions. I'll quote it here to keep this answer self-contained:

Theorem A: Suppose $H, K < G$ and $1^G_H = 1^G_K$ with $[G : H] = p^2$. Then either $H = K^g$ for some $g \in G$ or $p^\varepsilon =(q^n- l)/(q-1)$ with $\varepsilon \leq 2$, $n \geq 3$, and $q$ a prime power. Furthermore, for any such $p$, there is a group $G$ with $1^G_H = 1^G_K$ for nonconjugate subgroups $H$ and $K$ with $[G : H] = p^2$.

Gavioli even mentions that linked subgroups need not even be isomorphic. For example, in Riemannian Coverings and Isospectral Manifolds by Toshikazu Sunada, article which is cited by Gavioli, pairs of non-isomorphic linked subgroups of the Galois group of a Galois extension of $\mathbb{Q}$ are used in order to create isospectral yet non-isometric Riemannian manifolds if I understand correctly.