Luckly, I just found that I can answer this question by myself now, and the answer is 'Yes', the base-dependent symmetry operators become the *same* in the physical subspace, here is the proof (The notations used here are the same as those in Two puzzles on the Projective Symmetry Group(PSG)?):

Let $A$ be the symmetry operator(e.g., lattice translation, rotation, and parity symmetries, and time-reversal symmetry). First of all, $A$ should make sense in the physical subspace, in the sense that if $\phi$ is a physical state, then $A\phi$ should also be a physical state, this is true due to the fact $[P,A]=0$. Secondly, after a gauge rotation $\psi_i\rightarrow\widetilde{\psi_i}=R\psi_iR^{-1}=G_i\psi_i$, the symmetry operator $A$ defined in $\psi_i$ basis would changes to $\widetilde{A}=RAR^{-1}$ defined in $\widetilde{\psi_i}$ basis, now use the identity $PR=RP=P$ in Two puzzles on the Projective Symmetry Group(PSG)?, it's easy to show that $\widetilde{A}P=AP$, and hence for any physical state $\phi$, we have $\widetilde{A}\phi=A\phi$, which means that the symmetry operator $A$ is well defined in the physical subspace.

Note that $R$ is the local $SU(2)$ **gauge rotation** instead of **spin rotation**, and in the above proof we have used $[P,A]=[P,\widetilde{A}]=0$.

**Remark:** The *spin-rotation* symmetry operator is a little special in the sense that it is *basis independent* (This is obvious due to the SU(2) gauge structure of Schwinger-fermion representation).

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy