$\DeclareMathOperator\Aut{Aut}$Let $G$ be a (simply) connected Lie group whose semisimple part has no compact factors, and let $\Gamma $ be a lattice (uniform?) in $G$. Is it true that the stabilizer $\operatorname{Stab}_{\Aut(G)} (\Gamma)$ of $\Gamma$ in the group of automorphisms $\Aut(G)$ is always discrete? In some special cases this is true (R. Mosak, M. Moskowitz. Stabilizers of Lattices in Lie Groups, Journal of Lie Theory Volume 4 (1994) 1–16).
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$\begingroup$ You have to assume that $G$ has discrete center, otherwise the claim is clearly false. For semisimple groups the claim holds due to Zariski density of lattices. $\endgroup$– Moishe KohanCommented Jul 10 at 13:27

$\begingroup$ We are talking about arbitrary Lie groups. For Abelian Lie groups, the statement I indicated is true (we are not talking about the normalizer of a lattice in a Lie group, but about the stabilizer of this lattice in an automorphism group of this Lie group!). For semisimple Lie groups, the statement I indicated is almost obvious. $\endgroup$– Vladimir47Commented Jul 10 at 15:08

$\begingroup$ Oh, I see, I misread the question. $\endgroup$– Moishe KohanCommented Jul 10 at 15:24
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It turned out that the statement I indicated is not true in the general case! A counterexample is the follows.
The Lie group $G$ is the universal covering for the group $E_2$ of orientationpreserving motion of a twodimensional Euclidean plane. This Lie group is threedimensional, solvable and has the form ${\bf R} \cdot {\bf R}^2 $. The lattice $\Gamma = {\bf Z} \cdot {\bf Z}^2$. Here the stabilizer is isomorphic to ${\bf R}^2$ and therefore nondiscrete.

$\begingroup$ Ah, also $\mathbf{Z}^n$ in $\mathrm{Isom}^+(\mathbf{R}^n)=\mathbf{R}^n\rtimes\mathrm{SO}(n)$ works, for the same reason ($n\ge 2$). $\endgroup$– YCorCommented Jul 15 at 9:53