Non-conjugate subgroups that are conjugate in complexification In trying to come up with a counter-example in my line of research, I would like to find an example as follows:
$G$ is a semisimple Lie group with complexification $G^{\mathbb{C}}$. $H_1, H_2 \subseteq G$ are subgroups that are not conjugate as subgroups of $G$ but are conjugate as subgroups of $G^{\mathbb{C}}$.
Here is why I suspect this might be possible: we can certainly find pairs of elements that are conjugate in $G^{\mathbb{C}}$ but not in $G$. For instance, with $G = \text{SL}(2, \mathbb{R})$, $G^{\mathbb{C}} = \text{SL}(2,\mathbb{C})$ and $x = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $y = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ then $x$ and $y$ are conjugates in $G^{\mathbb{C}}$ as $x$ is simply the Jordan canonical form of $y$ but they are not conjugate in $G$ as any $z \in \text{GL}(2, \mathbb{R})$ with $z^{-1} y z = x$ has negative determinant.
Unfortunately the example $G = \text{SL}(2, \mathbb{R})$ does not work at the level of subgroups: the two $2$-dimensional subgroups of $\text{SL}(2, \mathbb{R})$ are not equivalent in $\text{SL}(2, \mathbb{C})$ and the one-parameter subgroups are classified by the traces of their elements and are therefore equivalent in $\text{SL}(2, \mathbb{C})$ if and only if they are in $\text{SL}(2, \mathbb{R})$.
 A: Define $M_s=\begin{pmatrix}0 & 0 & 1 & 0\\ 0 & 0 & 0 & s\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$, and $G_s=\exp(\mathbf{R}M_s)\subset\mathrm{SL}_4(\mathbf{R})$. Then $G_1$ and $G_{-1}$ are not conjugate in $\mathrm{SL}_4(\mathbf{R})$ while they are conjugate in $\mathrm{SL}_4(\mathbf{C})$.

$M_1$ and $M_{-1}$ are conjugate in $\mathrm{GL}_4$. Since over $\mathbf{C}$ any two conjugate matrices are conjugate by a matrix of determinant one (multiply the conjugating matrix by a suitable scalar). So $M_1$ and $M_{-1}$ are conjugate in $\mathrm{SL}_4(\mathbf{C})$.
 
Argument for non-conjugation over $\mathbf{R}$: one has to check that $M_1$ is conjugate to no scalar multiple of $M_{-1}$:
Let $M_E$ be the block matrix $\begin{pmatrix}0& E\\ 0 & 0\end{pmatrix}$, for $E$ some square matrix. Suppose $E,F$ invertible. If $UM_EU^{-1}=M_F$, $U$ preserves the common kernel of $E$ and $F$, is block upper-triangular, say of the form $\begin{pmatrix} A & C \\ 0 & B\end{pmatrix}$. The conjugacy then writes as $AE=FB$. So $\det(A)\det(E)=\det(F)\det(B)$. If $U$ has determinant 1, then $\det(B)=\det(A)^{-1}$. So $\det(F)\det(E)^{-1}=\det(A)^2$ is a square. In the present case, $E=I_2$ and $F=\lambda\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$ has negative determinant so $\det(F)\det(E)^{-1}=-\lambda^2$.
