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Example of almost-conjugate subgroups
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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 $\operatorname{diag}(1, 1, -1, -1)$.

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 $i \in \{1, 2, 3\}$.

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 $\operatorname{diag}(1, 1, -1, -1)$.

Too much halving
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LSpice
LSpice
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  • 69

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 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$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 $i \in \{1, 2, 3\}$.

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

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 $i \in \{1, 2, 3\}$.

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LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

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