Let $G$ be a complex connected reductive group, and let $P \subseteq G$ be a parabolic subgroup. My question is the following: if $g$ and $h$ are elements of $P$ which are conjugate *as* elements of $G$, are they necessarily also conjugate in $P$?

**Edit**: In order to rule out the cases pointed out in the comments, I'm refining my question.
Let $U \subset P$ be the unipotent radical, and $L = P/U$ the reductive quotient. Let $g$ and $h$ be elements of $P$, which project to the same element in $L$, and which are conjugate as elements of $G$. Are they then necessarily also conjugate in $P$?

**Further Edit:**
If $g$ and $h$ are elements of $P$ which project to the same element of $L$, and if $g$ and $h$ both lie in *some* (but not necessarily the same) Levi subgroup, then they are conjugate in $P$. This is because all Levi subgroups are conjugate by an element of $U$, and so we can conjugate $g$ and $h$ into the same Levi (without changing their projection), and then they must be equal. This implies that without loss of generality, we can assume that $g$ and $h$ have the same semisimple component.

Now there are counter-examples if we do not require $h$ and $g$ to lie in some Levi. For example, let $G = GL(3,\mathbb{C})$, and $P$ to be the subgroup of upper triangular matrices. Then the following two elements are conjugate in $G$ but not in $P$, and both project to the same element of the Levi:

$\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \ \ \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.

To rule out this case, we should then also require that $h$ lies in some Levi. The question is then whether the conjugate element $g$ must also lie in some (possibly different) Levi. By the above discussion, this would imply that $h$ and $g$ are conjugate in $P$.

anytwo elements? $\endgroup$4more comments