(This is a slightly reformatted and clarified version of [my question from math.SE](https://math.stackexchange.com/questions/671422/can-haar-measure-fail-to-be-bi-invariant-without-conjugation-shrinking-a-set/671452#comment1411591_671452), since I believe
  

the answer there is wrong and its poster has not responded to my comment in over two weeks.)
  
  

Let $\: \langle \hspace{-0.02 in}G,\hspace{-0.04 in}\cdot,\hspace{-0.04 in}\mathcal{T}\hspace{.02 in}\rangle \:$ be a locally compact Hausdorff topological group, and $\mu$
  

be a [left Haar measure](http://en.wikipedia.org/wiki/Haar_measure) on $\: \langle \hspace{-0.02 in}G,\hspace{-0.04 in}\cdot,\hspace{-0.04 in}\mathcal{T}\hspace{.02 in}\rangle$. Suppose that $\mu$ is *not* right-invariant. 



> 
> Does it follow that there is a Borel subset $B$ of $G$ and an element
>    
> 
>  $g$ of $G$ such that $\:\operatorname{closure}\hspace{.015 in}(B)\:$ is compact and for $C$ given by
>  $$C= \{ \; h\in G \:\: : \:\: g^{\hspace{-0.02 in}-1} h\hspace{.02 in}g \: \in \: B \; \}$$
>  which satisfies 
>  $\: C\subset B \;\;$ and $\;\; \operatorname{interior}\hspace{.02 in}(B\hspace{-0.04 in}-\hspace{-0.04 in}C\hspace{.02 in}) \neq \emptyset$ ?
> 
> 
> 


(In other words, is there a topological group such that the Haar measures are
  

not bi-invariant but there is no "single-conjugacy" explanation for that fact?)



Can Haar measure fail to be bi-invariant without conjugation shrinking a set?
I restate your question: *is it true that for every non-unimodular locally compact group $G$, there exists a Borel subset $B$ with compact closure and $g\in G$ such that $gBg^{-1}\subset B$ and $B\smallsetminus gBg^{-1}$ has non-empty interior?*


The answer is no: here is a simple and typical counterexample. Fix $c>1$ and consider the semidirect product $\mathrm{SOL}\_c=\mathbf{R}^2\rtimes\mathbf{R}$, where the action of $t\in\mathbf{R}$ is given by $t\cdot (x,y)=(e^tx,e^{-ct}y)$. This group is not unimodular (while $\mathrm{SOL}\_1$ is unimodular). Denote by $p$ the projection $\mathrm{SOL}\_c\to\mathbf{R}$ modulo the normal $\mathbf{R}^2$.


Let us check that your condition fails; assume we have $B$, $g$ satisfying the conditions. Then there exists $t\_0$ such that $B'=p^{-1}(\{t\_0\})\cap B$ contains at least 2 elements. Since $p^{-1}(\{t\_0\})$ is invariant by conjugation, we have $gB'g^{-1}\subset B'$. If $p(g)\neq 0$, then the action of $g$ on $p^{-1}(\{t\_0\})$ is an affine transformation whose linear part has two eigenvalues distinct from 1, and thus has a unique fixed point; it is actually conjugate to the linear map $\mathrm{diag}(e^t,e^{-ct})$ for some $t$ and thus the positive orbit of any non-fixed point is unbounded, contradicting $gB'g^{-1}\subset B'$ ($B'$ being bounded and having at least 2 points). Hence $p(g)=0$. Now we see that the conjugation by $g$ acts as a nontrivial translation on each $p^{-1}(\{t\})$ for $t\neq 0$. We deduce that $B$ is contained in $p^{-1}(\{0\})$, contradicting that it has non-empty interior. 


