 Solution

The original symmetry flips every $X$-basis bit, $s_i\mapsto s_i+1$, without changing any domain wall. Hencewith the present normalization. On the dual side, periodic domain walls obeyso the original global symmetry becomes a constraint on the dual symmetry sector.

Periodic versus antiperiodic boundary conditions determine whether the product of dual domain walls is $+1$ or $-1$. Conversely, the original even and odd $G$ sectors correspond to choices of dual boundary twist. Keeping all sectors therefore requires summing over both symmetry charges and both boundary conditions; within each matched sector the intertwiner is invertible up to normalization.

If an operator $O$ is symmetric, $[O,G]=0$, it descends consistently to a dual operator satisfying $DO=\tilde{O}D$. If it is nonsymmetric, it changes the global charge and cannot be represented by a local operator within one fixed dual boundary sector; its dual either changes the twist, acquires a disorder string, or is annihilated by the projected intertwiner.