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 $\text{SL}(2, \mathbb{R})$.



Non-conjugate subgroups that are conjugate in complexification
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 $\mathrm{SL}\_4(\mathbf{C})$.




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$M\_1$ 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 $\det(F)\det(E)^{-1}=-\lambda^2$.


