Let $\mathbb{G}$ be a connected reductive algebraic group $\mathbb{F}_q$-defined over $\bar{\mathbb{F}_q}$, and let $x\in\mathbb{G}$ be a semisimple element inside an $\mathbb{F}_q$-defined maximal torus $\mathbb{T}$. Let $\Phi$ be the root system of $\mathbb{G}$ wrt $\mathbb{T}$, and put $$\Sigma=\Phi(g)=\lbrace\alpha\in \Phi:\alpha(g)=1\rbrace.$$

There is a very clean criterion by Deriztiois for which closed subsystems $\Sigma\le \Phi$ can occur as $\Phi(g)$ for some $g\in \mathbb{G}$; namely, these are precisely the subsystems $\Sigma\le \Phi$ which admit a basis which is a subset of the set of affine simple roots of $\Phi$, and all such subsystems occur for some $g\in \mathbb{T}$. Furthermore, if $g\in\mathbb{F}_q$, then $\Sigma$ is stable under the action of the Frobenius map associated with the $\mathbb{F}_q$-structure on $\mathbb{G}$.

Question Given a subsystem $\Sigma\le \Phi$ as in the last paragraph, stable under the Frobenius map, does there exist $g\in \mathbb{T}(\mathbb{F}_p)$ for which $\Sigma=\Phi(g)$? Do there exist counterexamples for this?

What I know by now: If $\Sigma=\Phi(g)$ is a pseudo-Levi subsystem, meaning that it has a basis of simple elements of $\Phi$ then one can always take $g$ to be $\mathbb{F}_p$-rational. To show this, one can compute the dimension of the subgroups $$\mathbb{T}_{\Sigma'}=\bigcap_{\alpha\in \Sigma'}\ker(\alpha)$$ for all $\Sigma\le \Sigma'\le \Phi$, and verify that $$\dim\mathbb{T}_{\Sigma'}\le \mathrm{rk}(\Phi)-\mathrm{rk}(\Sigma),$$ with equality iff $\Sigma'=\Sigma$, and, consequently, deduce that $\mathbb{T}_{\Sigma}^\circ\setminus(\bigcup_{\Sigma<\Sigma'}\mathbb{T}_{\Sigma'})$ is irreducible of dimension $\mathrm{rk}(\Phi)-\mathrm{rk}(\Sigma)$, and therefore admits an $\mathbb{F}_q$-rational point.