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Let $M$ be a rank-3 matrix, I am interested in searching all the group elements $g \in$ SU(3) Lie group, such that,

$$ g^T M g =M. $$

Example 1. Let $$ M[1]= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right),$$ then we can find that there is at least a subgroup $g \in SU(2) \subset SU(3)$ that makes the $M[1]$ satisfies $$ g^T M[1] g =M[1], $$ the $$ g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right) =\cos(\frac{\theta}{2})+i \sum_{k=1}^{3} t_k \sigma_k\sin(\frac{\theta}{2})$$

\begin{align} \sigma_1 = \sigma_x = \begin{pmatrix} 0&1 & 0\\ 1&0 & 0\\ 0&0 & 0 \end{pmatrix}, \sigma_2 = \sigma_y = \begin{pmatrix} 0&-i& 0\\ i&0& 0\\ 0 & 0& 0 \end{pmatrix}, \sigma_3 = \sigma_z = \begin{pmatrix} 1&0& 0\\ 0&-1& 0\\ 0 & 0 & 0 \end{pmatrix} \,. \end{align} Notice that any group element on $SU(2)$ can be parametrized by some $\theta$ and $(t_1,t_2,t_3)$. Also $\theta$ has a periodicity $[0,4 \pi)$.

Example 2. Let $$ M[1]= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), M[2]= \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right), M[3]=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),$$

Can we find some subgroup $g \in G \subset$ SU(3) Lie group? such that $$ g^T \{M[1], M[2], M[3]\} g =\{M[1], M[2], M[3]\} $$ This means that $g^TM[a]g=M[b]$ which may transform $a$ to a different value $b$. But overall the full set $ \{M[1], M[2], M[3]\}$ is invariant under the transformation?

Example 3. . Let $$ M[1]= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), M[2]= \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right), M[3]=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),$$ $$ M[4]= -\left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), M[5]= -\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right), M[6]=-\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),$$

Can we find some subgroup $g \in G \subset$ SU(3) Lie group? such that $$ g^T \{M[1], M[2], M[3],M[4], M[5], M[6]\} g =\{M[1], M[2], M[3],M[4], M[5], M[6]\} $$ This means that $g^TM[a]g=M[b]$ which may transform $a$ to a different value $b$. But overall the full set $ \{M[1], M[2], M[3],M[4], M[5], M[6]\}$ is invariant under the transformation?

How can we write a Mathematica .nb file to solve this?

(Of course, there is always a trivial solution the identity $g=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$), but what are other solutions?

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closed as off-topic by Daniel Lichtblau, Henrik Schumacher, m_goldberg, José Antonio Díaz Navas, Edmund Apr 10 '18 at 0:01

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