As we know, under the symmetry operation $U$, the operator $\hat A$ and the state $|\alpha \rangle $act as $$\hat A \longrightarrow U\hat A U^{\dagger}$$ $$|\alpha \rangle \longrightarrow U|\alpha\rangle$$ However, for the second quantization operator, i.e. $c^{\dagger}$ or $b^{\dagger}$, sometimes they transform like a state. For example, suppose we rotate the local spin basis(cf. Fradkin Chapter 2.2): $$c'_\sigma(r)=U_{\sigma\sigma'}c_{\sigma'}(r)$$ where $U$ is a $2 \times 2 $ SU($2$) matrix.
Or, under U($1$) symmetry transform: $$c'_\sigma(r)=e^{i\theta}c_{\sigma}(r)$$ Both of above case implies that the second quantization operator act like a state under symmetry transform.
However, under the transform between the Schrödinger picture and Heisenberg picture: $$c_{(H)}=e^{i\hat Ht} c_{(s)} e^{-i\hat Ht}$$ in this case, the second quantization operator act like a operator under transform. And I know that the above boson/fermion operator sometimes are just the combination of the "normal" operator, e.g. for harmonic oscillator $\hat a^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}} (\hat x+\frac{i}{m\omega}\hat p)$. So I think the second quantization operator should follow the transform style of "normal" operator naturally.
In summary, I wonder that what kind of style of symmetry transformation on the second quantization operator, i.e. operator or state? And are there some mistakes about my argument above? Thanks!