I recently became interested in the following question: Given a Lie algebra $\mathfrak{g}$, define two elements $x,y\in\mathfrak{g}$ to be equivalent if there exists an automorphism $\phi\in\operatorname{Aut}(\mathfrak{g})$ such that $\phi(x)=y$. Are there particular cases where one can understand this set? Any reference to results in this direction is appreciated. Thanks!
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$\begingroup$ So you are interested in the orbits of the action of $\mathrm{Aut}({\mathfrak g)}$ on $L$. When ${\mathfrak g}$ is abelian you just have 2 orbits but, of course, this is only a trivial case. $\endgroup$– Salvatore SicilianoCommented Sep 21, 2022 at 16:35
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$\begingroup$ For gl(n) there is the Noether-Skolem theorem, so then you are thrown back to classifying conjugy classes. This is linear algebra then. $\endgroup$– user473423Commented Sep 21, 2022 at 17:10
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$\begingroup$ For semisimple Lie algebras you can link orbits of nilpotent elements to orbits of parabolic subalgebras (I think this may be orbits of $\mathrm{Inn}(\mathfrak{g})$ which in this case is the connected component of the identity of $\mathrm{Aut}(\mathfrak{g})$). Key words are Bala-Carter theory, Jacobson-Morozov parabolics and Richardson orbits. $\endgroup$– CallumCommented Sep 22, 2022 at 7:20
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$\begingroup$ Thanks for all the comments and references! $\endgroup$– Joakim ArnlindCommented Sep 23, 2022 at 5:21
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