Is unimodular stable under local isomorphisms?

Question

Let $G$ and $H$ be locally compact groups. Suppose that $G$ and $H$ are locally isomorphic. If $G$ is unimodular, is it true that $H$ is unimodular ?

Two topological groups $G$ and $H$ are said locally isomorphic if there exists open neighborhoods $V_G$, $V_H$ of $e_G$ and $e_H$ and a homeomorphism $f:V_G \to V_H$ such that for all $x,y \in V_G$ such $xy \in V_G$ we have $f(xy)=f(x)f(y)$ and similarly for $f^{-1}$.

Answer 1

There is a counterexample in Dieudonné II Exercise 3 page 262 chapter XIV 3.

Answer 2

For $s\in\mathbf{R}^*$ consider the action of $\mathbf{Z}$ on $\mathbf{R}$ by $n\cdot x=s^nx$. Let $G_s$ be the semidirect product $\mathbf{R}\rtimes\mathbf{Z}$ given by this action (for $s=1$ it's the direct product).

Then for $s\neq\pm 1$ this is not unimodular, while for $s=1$ it is. While they're all locally isomorphic (to $\mathbf{R}$).

[Unimodularity is a property of the action of $G$ (acting globally) around $1$ (locally acted on)].

On the other hand for virtually connected Lie groups it is indeed a local isomorphism invariant, since then it holds iff the Lie algebra is unimodular (in the sense that the adjoint action is by trace zero elements).