Consider a Lie group $G$ and a Hilbert space $\mathbb{V}$ equipped with a dot product $~~\langle~,~\rangle$.
Let ${^G{\cal{L}}}$ be a space of square-integrable functions mapping $G$ to $\mathbb{V}$: $$ {^G{\cal{L}}}=\left\{ f:~~ G\longrightarrow\mathbb{V}\qquad\Big{|}\quad \int dg \langle f(g), f(g)\rangle < \infty \right\}. $$
We can map each $ f$ to a function of the inverse argument: $$ f(x)\longmapsto\varphi(x)\equiv f(x^{-1})~,\quad x\in G \qquad\qquad\qquad\qquad (*) $$ or, in short: $$ f\longmapsto\varphi\equiv f\circ\hat{\zeta}~, $$ where $$ \hat{\zeta} x\,=\,x^{-1} $$ is the inversion operation on $G$.
Doing this for all $f$, we obtain a new space of functions: $$ {\cal{L}}^{G}=\left\{~\varphi:~G\longrightarrow{\mathbb{V}}~~\Big{|}~~\varphi= f\circ\hat{\zeta}~,~~f\in{^G{\cal{L}}} \right\}~. \qquad\qquad\qquad (**) $$
To examine if $~^{G}{\cal{L}}$ and ${\cal{L}}^{G}$ are copies of the same functional space, check the square-integrability of the new functions: $$ \int dg \langle \varphi(g),\varphi(g)\rangle=\int d(g^{-1}) \langle \varphi(g^{-1}) , \varphi(g^{-1})\rangle= \int d(g^{-1}) \langle f(g),f(g)\rangle $$ For unimodular groups, the measure is invariant and $d(g^{-1}) = dg $, wherefrom $$ \int dg \langle \varphi(g),\varphi(g)\rangle= \int dg \langle f(g),f(g)\rangle < \infty~, $$ so the spaces $^{G}{\cal{L}}$ and ${\cal{L}}^{G}$ coincide.
For what nonunimodular groups would this outcome stay valid?
Say, for compact ones? For locally compact ones?
PS.
I understand that, by Haar's theorem, if a group is locally compact (i.e. if its identity element has a compact neighborhood), then it admits a unique left-invariant and a unique right-invariant measure (unique -- up to multiplication by a positive constant). Under the transformation $g\to g^{-1}$, a left-invariant measure transforms into a right-invariant one, and vice versa. Can this fact be somehow employed here?