Extending Wigner's Classification with Gauge Symmetry In Wigner's Classification, as far as I understand, one uses unitary irreps. of the Poincare group for treating an elementary particle, since then mass $m$ and helicity $h$ emerge naturally as properties of the particle that are invariant of any given reference frame.
Now, if this Poincare symmetry, postulated from special relativity led to a coarse classification of particles, does or can postulated gauge symmetry (in the according theories) play a similar role? I.e. can it give rise to conserved physical quantities, with which to identify and distinguish particles? Not like conserved current per se but for example the value of some charge $e$ in and of itself.
 A: Instead of gauge symmetry, a perhaps more meaningful (and related) question is how internal global symmetry enters the classification. Assuming the symmetry group is compact (compact Lie group or finite group), then the particles are also labeled by finite-dimensional irreducible representation of the internal symmetry group. Charge e is an example when the group is U(1).
A: Gauge invariance is not physical (aka global) symmetry, but is a redundancy you introduced for expressing your theory in the language of Langragian/Hamiltonian formalism. Gauge transformation although leaves the action and equations of motion invariant under local transformations of the fields, is an "do=nothing" transformation on quantum states and quantum observables. In other words, physical states in quantum field theory form a trivial representation of the gauge group.
Take $U(1)$-gauge theory as an example. It is well-known that the classical Gauss law $\nabla\cdot\vec{E}=\rho$ is promoted as an operator constraints on physical states, i.e $$(\nabla\cdot\vec{E}-\rho)|\Psi\rangle=0, \tag{1}$$
where $|\Psi\rangle$ is a physical state. What can you see from the above identity? I see the following thing:

Do an expotential map of the operator $(\nabla\cdot\vec{E}-\rho)$, i.e one defines the operator $$U\equiv\exp\left\{i\int d^{3}\mathbf{x}\left(\nabla\cdot\vec{E}(\mathbf{x})-\rho(\mathbf{x})\right)\lambda(\mathbf{x})\right\}.$$

Apparently, $U$ is an $U(1)$-valued operator. Then, equation (1) says that for any physical state $|\Psi\rangle$, the identity $$U|\Psi\rangle=|\Psi\rangle \tag{2}$$
must hold! In other words, all physical states must be invariant under the above $U(1)$-gauge transformation.
Moreover, in temporal gauge $A^{0}=0$, one has the equal-time canonical commutation relation $$[A_{i}(\mathbf{x}),E_{j}(\mathbf{y})]=\delta_{ij}\delta(\mathbf{x}-\mathbf{y}).$$
Then, equation (2) implies that $\delta A_{i}(\mathbf{x})=\partial_{i}\lambda(\mathbf{x})$, which you should recognize immediately as the infinitesimal $U(1)$-gauge transformation $A\rightarrow A+d\lambda$. For the same reason, gauge transformations should also leave physical observables unchanged.
On the other hand, global symmetries form unitary representations of the global symmetry group. Such transformations transform a physical state $|\Psi\rangle$ into some other physical state $|\Psi^{\prime}\rangle$. In QED, there is a $U(1)$-gauge redundancy as explained above, but there's also a global $U(1)$-symmetry in the fermionic/scalar sector, which leads to the conservation of the Noether charge known as the electric charge. The electric charge you obtained from Noether's theorem in QED is gauge invariant, and thus is a physical observable.
