I am considering the following question.



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> **Question 1.** Let $G$ be a reductive algebraic group over $\mathbb{R}$, can we find finitely many reductive $\mathbb{R}$-subgroups $H\_1,...,H\_m$ such that for any reductive $\mathbb{R}$-subgroup $H\subset G$, there is a $g\in G(\mathbb{R})$ such that $H=gH\_ig^{-1}$ for some $i$?
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We know that if one replace "reductive" in the above question to "semi-simple", then we get a positive answer, see [this question](https://mathoverflow.net/questions/237556/a-finiteness-property-for-semi-simple-algebraic-groups?noredirect=1&lq=1). 


Unfortunately, the answer to my **Question 1** is definitely "No". For example, take $G=\mathbb{G}\_{m,\mathbb{R}}^2$, and for each $i\in \mathbb{N}$, define the subtorus $H\_i=\{(x,x^i);x\in \mathbb{R}^\*\}$, then these $H\_i$ are not conjugate to each other. 


The reason why we have the above counter-example is that in a torus, conjugation is a trivial action. So in order to get non-trivial conjugations, I am trying to enlarge the ambient algebraic group. I modify my **Question 1** as follows.



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> **Question 2.** Let $G$ be a reductive algebraic group over $\mathbb{R}$, can we find a faithful representation $\rho: G\rightarrow \mathrm{GL}\_{n,\mathbb{R}}$ and finitely many reductive $\mathbb{R}$-subgroups $H\_1,...,H\_m\subset \mathrm{GL}\_{n,\mathbb{R}}$ such that for any reductive $\mathbb{R}$-subgroup $H\subset G$, there is a $g\in \mathrm{GL}\_n(\mathbb{R})$ such that $H=gH\_ig^{-1}$ for some $i$?
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In my new question, the conjugation element $g$ can be chosen more flexibly. So I would like to know whether **Question 2** is true. Thanks very much!



About the conjugation of reductive subgroups
No. Again the 2-torus $T$ is a counterexample. Indeed, first an embedding into $\mathrm{GL}\_n$ is given by $n$ characters, hence $n$ pairs $(k\_1,m\_1)$,... $(k\_n,m\_n)$ of integers such that the corresponding matrix defines a surjective endomorphism $\mathbf{Z}^n\to\mathbf{Z}^2$. Namely it's conjugate to the embedding 
$$(t,u)\mapsto \mathrm{diag}(t^{k\_1}u^{m\_1},\dots,t^{k\_n}u^{m\_n}).$$
For $\ell$, let $T\_\ell$ be the subtorus of $T$ of those $(t,u)$ with $u=t^\ell$. Then it maps to $\mathrm{diag}(t^{k\_1+\ell m\_1},\dots,t^{k\_n+\ell m\_n})$ [note that for every $\ell$, $\mathrm{gcd}(k\_1+\ell m\_1,\dots,k\_n+\ell m\_n)=1$]. For $\ell\_1,\ell\_2$, the images of $T\_{\ell\_1}$ and $T\_{\ell\_2}$ are conjugate in $\mathrm{GL}\_n$ if and only if the $n$-tuple $(k\_1+\ell\_2 m\_1,\dots,k\_n+\ell\_2 m\_n)$ is a signed permutation of $(k\_1+\ell\_1 m\_1,\dots,k\_n+\ell\_1 m\_n)$. For each given $\ell\_1$ this can hold only for finitely many $\ell\_2$, and hence these fall into infinitely many conjugacy classes.


