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Let $G$ be a connected linear algebraic group (say, over an algebraically closed field) and let $H < G$ be a closed connected subgroup. Let $N_G(H)$ be the normalizer of $H$ in $G$, and assume that $N_G(H)=H$. Does there necessarily exist a single element $h \in H$ such that if $g \in G$ satisfies $g h g^{-1} \in H$, then $g \in H$?

In the setting I am in $H$ is solvable, but I doubt that is relevant to the question.

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    $\begingroup$ "Say, over an algebraically closed field" is not just an aside—it's essential! Otherwise, this does not work if $H$ is a maximal split torus in $G = \operatorname{SL}_2$ over $\mathbb F_2$. I do not know a general result of the sort you seek, but is it possible that you know more about $H$, for example, that it is a Borel subgroup of $G$? $\endgroup$
    – LSpice
    Commented Aug 24, 2023 at 1:59
  • $\begingroup$ If $H$ is reductive, you take a regular element $h$. Then $H$ is the Zariski closure of the group generated by all $H$-conjugates of $h$. Then the claim follows. $\endgroup$
    – user473423
    Commented Aug 24, 2023 at 14:20
  • $\begingroup$ @Echo, re, how does the claim follow? I don't see why $g h g^{-1} \in H$ implies that the $g$-conjugate of every $H$-conjugate of $h$ lies in $H$ (or maybe that wasn't your point). $\endgroup$
    – LSpice
    Commented Sep 12, 2023 at 2:59
  • $\begingroup$ Yes, you're right, I was too optimistic. If $H$ is a torus, it works, otherwise it doesn't. $\endgroup$
    – user473423
    Commented Sep 12, 2023 at 9:16

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