a matrix $Q \in GL(3,{\mathbb{R}})$ such that $Q^{-1} \Gamma_2 Q = \Gamma_1$. The explicit forms of $\Gamma_1$, $\Gamma_2$ and $Q$ are given in Table \[tet\_reps\_table\]. Note that $Q$ can be chosen to be orthogonal. Let us define $\hat{Q}:=I_3 \oplus Q \in O(6, {\mathbb{R}})$, where $I_3$ denotes the $3 \times 3$ identity matrix; then we have $$\label{t_2}
\overline{{\mathcal{T}}}:= \hat{Q}^{-1} \hat{{\mathcal{T}}} \hat{Q} = \Gamma_1 \oplus \Gamma_1.$$ We consider the set ${\mathcal{Z}}(\overline{{\mathcal{T}}},{\mathbb{R}}) \cap SO(6)$. Writing a matrix $N$ in this group as $$N = \left( \begin{array}{cc}
N_1 & N_2 \\
N_3 & N_4 
\end{array} \right),$$ where $N_i$ are $3 \times 3$ matrices, for $i =1, \ldots, 4$, we impose $N \overline{{\mathcal{T}}} = \overline{{\mathcal{T}}} N$, i.e. $N (\Gamma_1 \oplus \Gamma_1) = (\Gamma_1 \oplus \Gamma_1) N$. Using Schur’s Lemma and imposing orthogonality, we obtain $$N = N(\beta) = \left( \begin{array}{cc}
\cos(\beta)I_3 & -\sin(\beta)I_3 \\
\sin(\beta)I_3 & \cos(\beta)I_3
\end{array} \right),$$ where $\beta$ belongs to the unit circle $S^1$. Notice that $N(\alpha)N(\beta) = N(\alpha+\beta)$. Putting together and we obtain $${\mathcal{Z}}({\widetilde{\mathcal{T}}},{\mathbb{R}}) \cap SO(6) = \left\{ (R\hat{Q}) N(\beta) (R\hat{Q})^{-1} :   N(\beta) = \left( \begin{array}{cc}
\cos(\beta)I_3 & -\sin(\beta)I_3 \\
\sin(\beta)I_3 & \cos(\beta)I_3
\end{array} \right), \; \beta \in S^1 \right\}.$$ It follows that the group ${\mathcal{Z}}({\widetilde{\mathcal{T}}},{\mathbb{R}}) \cap SO(6)$ is isomorphic to 