Suppose we have a finite group $G$ with subgroup $H$, a representation $\rho_V$ of $H$ on a finite-dimensional vector space $V$, and an $H$-invariant inner product on $V$:
$$\forall x,y\in V, h\in H,\enspace \langle\rho_V(h)x, \rho_V(h)y\rangle = \langle x,y\rangle$$
We will write $V_I$ for the direct sum of $\lvert G:H \rvert$ copies of $V$: $$V_I = \oplus_{i=1}^{\lvert G:H \rvert} V$$ We define a map $L_i$ that lifts $V$ into the $i$th copy in the direct sum: $$L_i: V\to V_I\\ L_i v = 0 \oplus 0 \oplus ... \underbrace{v}_{i\text{th summand}} \oplus 0 + ...$$ We extend the inner product on $V$ to one on $V_I$:
$$ \langle \sum_i L_i x_i, \sum_j L_j y_j \rangle = \sum_i \langle x_i, y_i \rangle$$ From each left coset $K_i$ of $H$ in $G$ we pick an element $k_i$, so $K_i=k_i H$. We then have an induced representation $\rho_I$ of the group $G$ on $V_I$: $$\rho_I(g) \sum_i L_i v_i = \sum_i L_{j(g,i)} \rho_V(k_{j(g,i)}^{-1} g k_i) v_i$$
where $j(g,i)$ is the index of the coset $K_i$ to which $g k_i$ belongs.
Now, suppose we have an irreducible representation $\rho_W$ of $G$ on some finite vector space $W$. The Frobenius reciprocity theorem says that $Hom_H(W,V)$, the space of $H$-intertwiners from $W$ to $V$, i.e. the space of maps $S$ that satisfy:
$$S: W\to V\\ \forall h\in H,\enspace S \rho_W(h) = \rho_V(h) S$$ is isomorphic to the space $Hom_G(W,V_I)$ of $G$-intertwiners from $W$ to $V_I$, i.e. the space of maps $T$ that satisfy:
$$T: W\to V_I\\ \forall g\in G,\enspace T \rho_W(g) = \rho_I(g) T$$
Indeed, given an intertwiner $S\in Hom_H(W,V)$ we can easily construct an intertwiner $T_S\in Hom_G(W,V_I)$:
$$T_S w = \sum_i L_i S \rho_W(k_i^{-1}) w$$
As we vary $S$ over any basis of $Hom_H(W,V)$, the associated map $T_S$ will map $W$ into distinct subspaces of $V_I$, each of which is invariant under the action of $\rho_I$, and each of which has the property that the restriction of $\rho_I$ to that subspace is equivalent to $\rho_W$.
My question is: supposing the dimension of $Hom_H(W,V)$ is greater than 1, does there exist an "easy" strategy to choose a basis $\{S_1,S_2,...\}$ of $Hom_H(W,V)$ such that the subspaces $T_{S_1}(W), T_{S_2}(W), ...$ of $V_I$ will be mutually orthogonal?
By "easy", I mean something less computationally demanding than simply finding all the subspaces by means of an arbitrary basis for $Hom_H(W,V)$, and then decomposing their direct sum into orthogonal invariant subspaces. In other words, I am seeking to leverage the fact that $Hom_H(W,V)$ is a lower-dimensional space than the direct sum of the $T_{S_i}(W)$ in order to carry out a less demanding procedure to achieve the same result.
Edited to add: One easy way to get a basis of $Hom_H(W,V)$ is to take a sufficient number of linearly independent maps $S^{(0)}_i: W\to V$ and average over the subgroup $H$:
$$S_i = \frac{1}{|H|}\sum_{h\in H} \rho_V(h) S^{(0)}_i \rho_W(h^{-1})$$
So by starting with a basis of (non-intertwining) maps from $W$ to $V$, you can project as many as required into $Hom_H(W,V)$ to obtain a basis.
My (possibly naive) hope is that there might be some way of modifying this construction to lead directly to a basis with the property I'm seeking.