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Let $G$ be an affine algebraic group over $k=\bar{\mathbb{F}_{p}}$. Let $q$ be a power of $p$, and assume that $G$ is defined over $\mathbb{F}_q$. Let $\mathcal{T}$, be the collection of all maximal torus of $G$. Let $F$ be the Frobenius map from $G$ onto $G$.

Clearly $G$ acts on $\mathcal{T}$, by conjugation and this action is transitive. Further if we assume that $G$ is connected, by standard theory in "Finite groups of Lie type", $G^{F}$, acts on $\mathcal{T}^{F}$ by conjugation,where $G^{F}$ denote the set of $F$-rational points of $G$, which is the finite algebraic group associated to $G$, and $\mathcal{T}^{F}$, denote the collection of $F-$ stable maximal tori of $G$,that is,

$$\mathcal{T}^{F}=\{ T\in \mathcal{T} | F(T)=T \}. $$

We know, $G^F$ need not act transitively on $\mathcal{T}^F$. For each $T\in \mathcal{T}^F$, Let $T^F$, denote the $F-$ rational points in $T$. Now, let $$\mathcal{T}^{[F]}= \{ T^F | T\in \mathcal{T}^{F} \}.$$

It is clear that $G^F$ acts on $\mathcal{T}^{[F]}$, by conjugation. Also, if $T_1, T_2 \in \mathcal{T}^F$ are in same orbit under $G^F$ action then $T_1^{F},T_2^{F}$ are also in same orbit under $G^F$ action.

My question is Is it true other way around, that is, if $T_1^{F}, T_2^{F}$, are $G^F$-conjugate, then $T_1,T_2$ are $G^F$ conjugate. It seems to be that it is true. But, I couldn't come up with any proof.

I would appreciate any kind of help. Thank you!

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