My background about this argument:
a) Let's consider a classical field theory, where $\mathcal{L}(\phi(x),\partial_{\mu}\phi_(x))$ is the lagrangian density. A symmetry is a transformation $\phi(x) \rightarrow\phi'(x')=f(\phi(x))$ of the fields that leaves the lagrangian density $\mathcal{L}$ invariant (more general definitions are possible, but I hope this one is sufficient). Noether's theorem establishes that continous symmetry groups correspond to conserved currents and charges.
b) In the corresponding quantum field theory, fields are promoted to fields operators acting on the Hilbert space of states. In this case, symmetries are represented by probability conserving operators. According to Wigner's theorem, these are unitary operators (or antiunitary) on the Hilbert space of states. Conserved quantities should instead arise as operators that commute with the Hamiltonian. The latter is "imported" in the quantum field theory just by considering directly the classical expression, where field operators appear instead of classical fields.
Example:
Dirac free lagrangian density is: $$\mathcal{L}=\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$
Parity is a (discrete) symmetry of this lagrangian at a classical level, since ($'$ denotes parity-transformed quantities): $$\mathcal{L'}=\overline{\psi'}(i\gamma^\mu\partial_\mu'-m)\psi'=\overline{\gamma^0\psi}(i\gamma^\mu\partial_\mu'-m)\gamma^0\psi=\psi^+(i\gamma^\mu\partial_\mu'-m)\gamma^0\psi=\psi^+\gamma^0(i\gamma_\mu\partial^\mu-m)\psi=\mathcal{L}.$$
Questions:
a) Does the classical symmetry imply the existence of an operator that in some sense "represents" parity symmetry at a quantum level?
b) If yes, what is the connection between the operator and the classical parity transformation?
c) Does a conserved quantity at a quantum level (operator that commute with hamiltonian) arise in this case as a consequence of the existence of the parity symmetry?
d) Is it possible to generalize the considerations about the special simple case of parity?