Let $G$ be an adjoint algebraic group over $\mathbb{C}$, $\mathfrak{g}$ its Lie algebra.
Let $\rho:G\rightarrow GL(\mathfrak{g})$ be the adjoint representation. Let $g,g'\in G$ be two semisimple elements such that $\rho(g),\rho(g')$ are conjugated in $GL(\mathfrak{g})$, when do we have that $g$ and $g'$ are conjugated?


I am already interested in the case where $G=PGL\_n$.



conjugacy in adjoint representation
Here is a partial answer. Let first $G=PGL(n)$. Let $g$ have eigenvalues $\lambda\_1,\ldots,\lambda\_n$ (up to a scalar). Then $\rho(g)$ has eigenvalues $\lambda\_i\lambda\_j^{-1}$ with $i,j=1,\ldots,n$. This set invariant under $z\mapsto z^{-1}$. So $\rho(g)$ is always conjugate to $\rho(g^{-1})$. But for $n\ge3$ there exist $g$ which is not conjuguate to $g^{-1}$, e.g., $\mathrm{diag}(2,1,\ldots,1)$. So $G=PGL(n)$ does not work for $n\ge3$.


This argument can be generalized: Assume the Dynkin diagram has a non-trivial graph automorphism $\theta$. Then $\theta$ induces an outer automorphism $\theta\_G$ of $G$ and an automorphism $\theta\_{\rm ad}$ of $\mathfrak g$. Then $\rho(\theta\_G(g))=\theta\_{\rm ad}\rho(g)\theta\_{\rm ad}^{-1}$ is conjugate to $\rho(g)$ but there exist $g$ which are not conjugate to $\theta\_G(g)$ (otherwise $\theta\_G$ wouldn't be outer).


Conclusion: $G$ can't have outer automorphisms. I don't know what happens in these cases.


