A pair of non-conjugate subgroups: a simple proof

Question

$\DeclareMathOperator\SO{SO}$Set \begin{equation} \begin{aligned} \Gamma_1 &= \left\{ I_{6}, \; \gamma_1:= \left( \begin{smallmatrix} 0&1\\ 1&0 \\ &&0&1\\ &&1&0\\ &&&&1 \\ &&&&&1 \end{smallmatrix} \right), \; \gamma_2:= \left( \begin{smallmatrix} 0&1\\ 1&0 \\ &&1&&&\\ &&&1&&\\ &&&&0&1\\ &&&&1&0 \end{smallmatrix} \right), \; \gamma_3:= \left( \begin{smallmatrix} 1&&&\\ &1& \\ &&0&1&\\ &&1&0\\ &&&&0&1&\\ &&&&1&0 \end{smallmatrix} \right) \right\}, \\ % % % \Gamma_2 &= \left\{ I_{6}, \; \gamma_1':= \left( \begin{smallmatrix} 0&1\\ 1&0 \\ &&0&1\\ &&1&0\\ &&&&1\\ &&&&&1 \end{smallmatrix} \right), \; \gamma_2':= \left( \begin{smallmatrix} 0&&1&\\ &0&&1 \\ 1&&0&\\ &1&&0\\ &&&&1\\ &&&&&1 \end{smallmatrix} \right), \; \gamma_3':= \left( \begin{smallmatrix} 0&&&1\\ &0&1 \\ &1&0\\ 1&&&0\\ &&&&1\\ &&&&&1 \end{smallmatrix} \right) \right\}. \end{aligned} \end{equation}

I was able to prove that $\Gamma_1$ and $\Gamma_2$ are not conjugate in the special orthogonal group $SO(6)$, that is, there is no $a\in SO(6)$ such that $a\Gamma_1a^{-1}=\Gamma_2$. However, the proof is quite involved.

Do you have a simple argument to prove that $\Gamma_1$ and $\Gamma_2$ are not conjugate in $SO(6)$?

My interest in the pair of group is that they almost conjugate, that is, there is a bijection between the groups that preserves $SO(6)$-conjugacy classes. This follows easily since the eigenvalues of any non-trivial element are $+1$ with multiplicity $4$ and $-1$ with multiplicity $2$.

Answer 1

I think that the elements $g = \dfrac1{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^{\oplus3}$ and $h = \dfrac1 2\begin{pmatrix} 1 & 1 & 1 & 0 & 1 & 0 \\ -1 & 1 & -1 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & -1 & 0 \\ 1 & 1 & -1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{pmatrix}$ in $\operatorname{SO}(6)$ satisfy $g^{-1}\gamma_i g = h^{-1}\gamma_i'h$ for all $i \in \{1, 2, 3\}$.

If I understand correctly, one example of almost-conjugate subgroups is the pair of subgroups of $\operatorname{PGL}_4(\mathbb C)$ generated, first, by the images of $\gamma \mathrel{:=} \operatorname{diag}(1, -1, 1, -1)$ and $\operatorname{antidiag}(1, 1)^{\oplus2}$; and, second, by the images of $\gamma$ and $\operatorname{diag}(1, 1, -1, -1)$.