About the conjugation of reductive subgroups

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!

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.