I am considering the following question.

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 $i$?

We know that if one replace "reductive" in the above question to "semi-simple", then we get a positive answer, see this question.

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.

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 $i$?

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!