Let $H\trianglelefteq G$ be a normal closed inclusion of locally compact groups. Then $G$ acts by conjugation on $H$, thus on the one-dimensional space of Haar measures on $H$. For this action to be trivial, is it necessary for $G$ and $H$ to be unimodular? Is it sufficient? (We may as well assume, say, $G$ is second-countable for simplicity if it helps, or that it is a Lie group over a local field.)
It is evidently necessary that $H$ be unimodular. On the other hand, this is not sufficient. For example, take the inclusion $\mathbb{R}^n\trianglelefteq\mathbb{R}^n\rtimes\mathrm{GL}_n(\mathbb{R})$; the conjugation action on $\mathbb{R}^n$ manifests every matrix, many of which scale Lebesgue measure by a factor which is not $1$.
Edit: it is also sufficient that $H$ be compact or discrete, so it is not necessary that $G$ be unimodular. In the case $G=H\rtimes\mathrm{Aut}(H)_{\mathrm{disc}}$, I am asking how we can tell if every automorphism of $H$ scales Haar measure by $1$.
Further edit: If $H$ is open in $G$, then it does hold that unimodularity of $G$ and $H$ is necessary and sufficient.