Let $H$ be an open subgroup in a locally compact group $G$, $\iota:H\to G$ the embedding of $H$ into $G$, $\pi:H\to B(X)$ a unitary representation of $H$ in a Hilbert space $X$, and $\rho:G\to B(Y)$ the corresponding induced representation of $G$. When does there exist a (continuous in a proper sense) involutive homomorphism of algebras $\varphi:B(X)\to B(Y)$ such that 
$$
\varphi\circ\pi=\rho\circ\iota \quad ?
$$
This seems to be true at least when $H$ is finite and $\pi$ is irreducible. Is it possible that this is always true?



When is the induced representation factored through the initial one?
This is false in many cases where $G$ is finite. Let $\rho\circ \iota$ and $\pi$ denote the corresponding maps of group algebras $\mathbb{C}[G]$. The equation above can only hold if any element killed by $\pi$ must also be killed by $\rho\circ \iota$, that is $\mathrm{ker}(\pi)\subset \mathrm{ker}(\rho\circ \iota)$. If $\pi$ is irreducible, then $\mathrm{ker}(\pi)$ is the sum of all the other matrix algebra summands in the group algebra, so $\mathrm{ker}(\rho\circ \iota)$ can only contain $\mathrm{ker}(\pi)$ if $\rho\circ \iota$ is just a bunch of copies of the same irrep. Of course, this could happen, but it doesn't have to; for example, let $\pi$ be trivial, and $H$ any non-normal subgroup. The restriction of the permutation rep of $G/H$ to $H$ is non-trivial, so it can't factor through the trivial rep.


