I have seen the following statement made regarding the $\Bbb Z_2$ as a subgroup of $\operatorname{SO}(6)$: Viewed as a subgroup of $\operatorname{SO}(6)$, the $\Bbb Z_2$ generators are
$\Bbb Z_2 = \operatorname{diag}\{-1,-1,-1,-1,1,1\}$
$\Bbb Z_2 = \operatorname{diag}\{1,1,-1,-1,-1,-1\}$
I don't understand this, since, $\Bbb Z_2$ has 2 generators and in the 1D representation they are simply
$\Bbb Z_2 =\{-1,1\}$
What are these generators in the 6D picture (which one correspond to the identity 1, and which one correspond to the -1 element)?
Furthermore, how can one build a 6D representation of $\Bbb Z_2$ in $\operatorname{SO}(6)$ that look like those? Shouldn't they be simply
$\Bbb Z_2 = \operatorname{diag}\{-1,-1,-1,-1,-1,-1\}$ and $\Bbb Z_2 = \operatorname{diag}\{1,1,1,1,1,1\}$ ?
As @DerekHolt states in the comments . . .
Any diagonal matrix with entries from $\{-1,1\}$ and determinant $1$, excluding the identity, generates a subgroup of $\operatorname{SO}(6)$ isomorphic to $\Bbb Z_2$.