Let $G$ be a linear algebraic $\mathbb{Q}$-group, which is assumed to be connected, $\mathbb{Q}$-simple, and of adjoint type, such that the Lie group $G(\mathbb{R})$ has no compact factor defined over $\mathbb{Q}$. Let $\Gamma\subset G(\mathbb{Q})$ be a congruence subgroup. It is known, from the theory of Margulis, that $\Gamma\subset G(\mathbb{R})$ is Zariski dense. For convenience assume that $\Gamma\subset G(\mathbb{R})^+\cap G(\mathbb{Q})$ and that $\Gamma$ is torsion free. Note also that in this case, if one takes $X$ to be the non-compact symmetric domain associated to $G(\mathbb{R})^+$, then the quotient $X/\Gamma$ is a localy symmetric manifold of negative curvature (a typical example of hyperbolic manifold}.

I'd like to consider conjugates of linear $\mathbb{Q}$-subgroups of $G$ under $\Gamma$. More restrictively, let me take $H\subset G$ a connected semi-simple $\mathbb{Q}$-group such that $H(\mathbb{R})$ again has no compact factors defined over $\mathbb{Q}$. Then

(1) is the union $\bigcup_{g\in\Gamma}gHg^{-1}$ Zariski dense in $G$?

(2) if $\Gamma'$ is a finitely generated subgroup of $\Gamma$, and $H'$ be the Zariski closure of the subgroup of $G(\mathbb{Q})$ generated by $\bigcup_{g\in \Gamma'}gH(\mathbb{Q}) g^{-1}$, then how far is $\Gamma'$ from being an arithmetic subgroup of $H'$?

Thanks!