Let $\Gamma$ be a countable discrete group and $\lambda$ the left regular representation of $\Gamma$ on $l^2(\Gamma)$. Let $\rho:\Gamma\rightarrow U(H)$ be a unitary representation of $\Gamma$ on some separable infinite-dimensional Hilbert space $H$, and consider the representation $\lambda\otimes\rho$ of $\Gamma$ on $l^2(\Gamma)\otimes H$.

Now let $H_0$ be the same Hilbert space as $H$ but equipped with the trivial representation of $\Gamma$; then one can show that, as $\Gamma$-representations,

$$l^2(\Gamma)\otimes H\cong l^2(\Gamma)\otimes H_0,$$

where $\Gamma$ still acts by the left regular representation on the first factor on the right. For example one isomorphism is the map $\delta_g\otimes u\mapsto\delta_g\otimes(\rho(g)\cdot u)$, where $\delta_g:\Gamma\rightarrow\mathbb{C}$ is the function taking value $1$ at $g\in\Gamma$ and $0$ elsewhere.

I would like to know whether this can be generalised to the case of $G$ a Lie group instead of $\Gamma$. More specifically:

Question 1: Given a unitary representation $\rho:G\rightarrow U(H)$, is there an isomorphism $$L^2(G)\otimes H\cong L^2(G)\otimes H_0,$$ where $G$ acts on $L^2(G)$ by the left regular representation and $H_0$ is the same space as $H$ but with the trivial representation?

I'm not sure if the left-regular representation of $G$ on $L^2(G)$ is the right representation to use here, since the proof in the discrete case relies on the fact that the $\delta_g$ form a basis for $l^2(\Gamma)$, which doesn't seem to have an analogue in the continuous case. If this is hard to establish, I would also like to ask:

Question 2: Can one replace $L^2(G)$ by another representation in question 1 so that the conclusion is true?

Thanks!