Under what geometric conditions is the image of an irrep of $H$ isomorphic to the image of an irrep of the fundamental group of $G/H$?

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Let $G$ be a Lie group and $H$ a closed subgroup. Let $\pi_1(G/H)$ denote the fundamental group of the homogeneous space $G/H$. If $\lambda$ is an irrep of $H$, then under what conditions does there exist an irrep $\rho$ of $\pi_1(G/H)$ such that $$ H/\mathrm{ker}(\lambda) = \pi_1(G/H)/\mathrm{ker}(\rho). $$ In other words, under what conditions does there exist an irrep $\rho$ such that the image of $\lambda$ is isomorphic to the image of $\rho$? I am hoping for something geometric in nature (if possible).


If $G$ is simply connected and $H$ is discrete then it is known that $\pi_1(G/H) = H$ and so we can choose $\rho$ to be $\lambda$. However, if we drop the requirement that $H$ be discrete then more generally we have that $\pi_1(G/H) = \pi_0(H)$ (see this answer). It is less clear in this case when the above equation holds.


Context: This equation has come up regarding the "canonical" or "$H$-connection" one can place on the vector bundle $\mathrm{Ind}_H^G \lambda$ (that is, the "twisted vector bundle" underlying the induced representation). In particular $H/\mathrm{ker}(\lambda)$ is the "monodromy group" of the connection and $\pi_1(G/H)/\mathrm{ker}(\rho)$ seems to be related to classification of the flatness of the connection.