Is it possible that the scattering matrix is the minus identity matrix $S=-\mathbb{I}$? We know that if the scattering matrix is an identity matrix ($S=+\mathbb{I}$), it means that transmission is zero and there is full reflection.
My question:
Is it possible that the scattering matrix is the minus identity matrix ($S=-\mathbb{I}$)? Then, what does this imply for transmission and reflection in this case?
 A: I don't think that it means something, because transmission or reflection coefficients are defined as :
$$ T,R = |S_{ij}|^2 ,\tag 1$$
so under transformation of identity matrix:
$$ \begin{pmatrix}
    -1 & 0\\
    0 & -1
  \end{pmatrix}
$$
coefficients defined in (1) will have same values, i.e. they are invariant to sign transformation.
But for example, transposed identity matrix:
$$ S^T = \begin{pmatrix}
    0 & 1\\
    1 & 0
  \end{pmatrix}
$$
will have a reversed meaning - full transmission and zero reflectance.
A: I am not sure about you first statement. The S-matrix I know, when equal to identity, means nothing is happening. Hence transmission is one (full) and reflection is zero.
In principle there is nothing wrong with S-matrix being equal to $-\mathbb{1}$. The only requirement is that S-matrix is unitary. The simplest example for such S-matrix is a two-level spin$-1/2$ system evolving in time in magnetic field. If time duration $T$ is equal to a half of the period of Larmor precession, then $S(T)=-\mathbb{1}$.
More generally, $S=-\mathbb{1}$ just means that after the "active phase" of evolution is over, each basis state acquires phase $\pi$, without mixing with any other state. Therefore, just as in the $S=\mathbb{1}$-case, no reflection for $S=-\mathbb{1}$. Only "forward scattering".
