This question is straightforward. Is it possible for a unit length operator (vector, multivector, etc.), excepts for $-1$ itself that squares to $+1$ to have its hermitian conjugate equal to $-1$?, i.e, let's suppose we have $o^2=1$, is it possible to have $o^{*}=-o$ in in at least one representation?
My question comes by observing that for $1 \times 1$ matrix representation, the operator $i$ squares to $-1$ and conjugate to $-i$. The operator $1$, however, has the opposite behaviour.
For $2 \times 2$ matrix representation we can use Pauli matrices $\left( \sigma_i \right)$: $\sigma_i^2=1$, $\sigma_i^{*}=\sigma_i$.
In other words, does $o^{*}=o^2 o$ hold?
I found the solution for me.
Conjugation behaviour is done by definition based on "nice" properties that that definition have, like facilitating in the calculation of norms and inverses, so the answer may not be unique.
PS: For me Clifford conjugation worked.