Conjugacy classes in the automorphism group of a simple Lie algebra A lower bound of the number of conjugacy classes in the automorphism group of a simple Lie algebra $\mathfrak{s}$, of finite dimension over an arbitrary field $\mathbb{F}$, can be the size of the image of the automorphism group by the trace.
I think this map $\operatorname{tr}:\operatorname{Aut}(\mathfrak{s})\longrightarrow\mathbb{F}$ must be surjective. However, I have no clue. (For splitable Lie algebras, there is an explicit way).
 A: *

*If $\mathfrak{s}$ is $K$-anisotropic, where $K$ is a real or $p$-adic field (this is equivalent to $\mathfrak{s}$ not containing $\mathfrak{sl}_2$, and also to the corresponding group be compact), then the trace is bounded, hence non-surjective.


*Suppose that $\mathrm{Aut}(\mathfrak{s})^0$ is Zariski-dense in the underlying algebraic group (this is automatic if $K$ is perfect, by Rosenlicht's theorem). In this case, by Zariski density, if $\mathrm{Aut}(\mathfrak{s})^0$ has infinitely many traces (resp. characteristic polynomials), then the same holds over the algebraic closure. So we can reduce (in this case) to the algebraically closed case.


*In the algebraically closed case, if the number of characteristic polynomials of $\mathrm{Aut}(\mathfrak{s})^0$ is finite, it is (by connectedness) reduced to one, and hence it follows that every element in $\mathrm{Aut}(\mathfrak{s})^0$ is unipotent. In characteristic zero, this is not possible. I'm not sure about the modular case.
The same conclusion holds for the number of traces, although the characteristic zero is then used in a stronger way.


*If $\mathfrak{s}$ is the Lie algebra of a simple algebraic group, the restrictions on the characteristic should be dropped (still needing Rosenlicht), to infer that the number of characteristic polynomials is infinite.

