TL;DR:

Given representations $D,\Lambda$ of subgroups $K,Q$ of a Lie group $G$, is it true that every intertwining operator $T$ between the resulting induced representations of $G$ can be written
$$
(T\varphi)(g) = \int_G t(g^{-1} g')\varphi(g')\,dg'
\tag1
$$
for some function $t$ on $G$?

LONG VERSION:

I placed this question at the Mathematics StackExchange. Although I presumed that the question is simple if not trivial, no one was able to help me there. So I am now elevating it to the MathOverflow.

Consider the space $\,{\cal{L}}^G\,$ of all continuous functions $\,G\longrightarrow{\cal{L}}\,$ mapping a Lie group $\,G\,$ into a vector space $\,{\cal{L}}\,$.

Assume that $\,G\,$ has two proper subgroups: $$ K\,,~Q~<~G~~, $$ whose representations, $\,D(K)\,$ and $\,\Lambda(Q)\,$, are acting in $\,{\cal{L}}^G\,$.

Consider two subspaces of $\,{\cal{L}}^G\,$. One subspace,
$$
{\mbox{Map}}_K(G,\,{\cal{L}})\,=\,\left\{\,\varphi\,\right\}~~,
$$
comprises the vector functions $\,\varphi\,$ obeying the equivariance condition
$$
\varphi(g\, k)~=~D^{-1}(k)~\varphi(g)~,~~~k\,\in\, K~~.
$$

In this subspace, $\,D(K)\,$ is induced to a representation of $\,G\,$, denoted by
$$
U^{(D)}\,\equiv\,D(K)\,\uparrow\, G
$$
and implemented with
$$
U^{(D)}_g\,\varphi(g^{\,\prime})~=~\varphi(g^{-1}\, g^{\,\prime})~~.
$$

Another subspace, $$ {\mbox{Map}}_Q(G,\,{\cal{L}})\,=\,\left\{\,\psi\,\right\} $$ will comprise the functions $\,\psi\,$ satisfying $$ \psi(g\, q)~=~\Lambda^{-1}(q)~\psi(g)~,~~~q\,\in\, Q~~. $$ In this subspace, $\,\Lambda(Q)\,$ is induced to a representation of $\,G\,$, denoted by $$ U^{(\Lambda)}\,\equiv \,\Lambda(Q)\,\uparrow\, G $$ and implemented with $$ U^{(\Lambda)}_g\,\psi(g^{\,\prime})~=~\psi(g^{- 1}\, g^{\,\prime})~~. $$

While both $\,U^{(D)}\,$ and $\,U^{(\Lambda)}\,$ are realised via left translations, they are different representations, as they are acting in subspaces defined by different subsidiary conditions.

For convenience, we summarise this in the table: $$ \varphi(g\, k)=D^{-1}(k)\,\varphi(g)\,,~k\in K \quad \quad \psi(g\, q)=\Lambda^{-1}(q)\,\psi(g)\,,~q\in Q $$ $$ U^{(D)}\,\equiv\,D(K)\,\uparrow\, G \qquad \qquad \qquad U^{(\Lambda)}\,\equiv \,\Lambda(Q)\,\uparrow\, G $$ $$ U^{(D)}_g\,\varphi(g^{\,\prime})~=~\varphi(g^{-1}\, g^{\,\prime}) \quad \quad \qquad U^{(\Lambda)}_g\,\psi(g^{\,\prime})~=~\psi(g^{- 1}\, g^{\,\prime})~ $$

Our goal is to describe the space $\,\left[\, D(K)\,\uparrow\, G\,,~\Lambda(Q)\,\uparrow\, G \,\right]\,$ of the morphisms $\,\psi\,=\,\hat{T}\,\varphi\,$.

QUESTION:

How to prove that the most general form of a morphism is $$ \psi(g)~=~(\hat{T}\,\phi)(g)~=~\int_G t(g^{-1}\, g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}~~,\qquad\qquad\qquad(1) $$ where $\,dg\,$ is an invariant measure on $\,G\,$.

PS.

As an aside, I would mention that for the equivariance conditions to be satisfied the kernel must obey one more condition: $$ t(qgk) = \Lambda(q) t(g) D(k)~~,~~~q\in Q\,,~~g\in G\,,~~k\in K~~. $$ This, however, is the next theorem; and I don't want to go there until the basic property (1) is proven.