I am not having the best luck getting this, so any advice is appreciated. I apologize in advance if the question is too low-level. Let $G$ be a locally compact group $\sigma$-compact group, and $H\leq G$ a closed subgroup such that $G/H$ carries a finite invariant measure $\mu$. Let $(\pi,\cal H_\pi)$ be a unitary representation of $H$, and let $\rho={\rm Ind}_H^G(\pi)$, and call the Hilbert space it acts on by left translations ${\cal H}_\rho$. This space can be realized as the Hilbert space completion of the space of functions $G\to\cal H_\pi$ which are $H$-equivariant, with support compact modulo $H$, with respect to the inner product
$$
\langle f_1,f_2\rangle=\int_{G/H} \langle f_1(g),f_2(g)\rangle d\mu(gH).
$$
This is the standard simplest construction. My question is: **how can I define a bounded linear map $\Phi:{\cal H}_\pi\to{\cal H}_\rho$ which is an isometry and commutes with orthogonal projections onto subspaces of invariant vectors.**

What I tried is to let $\xi\in{\cal H}_\pi$ and define the function $f_\xi$ by $$f_\xi(x)=\begin{cases}\pi(x^{-1})\xi&\text{if}~x\in H,\\0&\text{otherwise.}\end{cases} $$ This defines a map $\Phi:\cal H_\pi\to\cal H_\rho$, $\xi\mapsto f_\xi$. But then, what I obtain is $$ \langle \Phi(\xi),\Phi(\eta)\rangle_{\cal H_\rho}=\mu(H)\langle \xi,\eta\rangle_{\cal H_\pi}, $$ with $\mu(H)$ the measure of the identity coset in $G/H$, since $f_\xi,\,f_\eta$ are non-trivial only on $H$. My map seemed fairly natural, is there something I am overlooking? Is there some twist to apply to this to make it work?