The gauge symmetry in classical pure Yang-Mills theory with a gauge field $A_{\mu}$ requires an action $S$ to be invariant under continuous transformations $$ A_{\mu}(g) \to g(A_{\mu} + i\partial_{\mu})g^{-1} $$ When we talk about quantized theory, we're dealing with the Hilbert space of rays $|\Psi (A_{\mu})\rangle$, which must be invariant under unitary transformation $U(g)$: $$ \tag 0 |\Psi(A_{\mu})\rangle \to U(g)|\Psi (A_{\mu})\rangle = |\Psi(A_{\mu})\rangle $$ Equivalently, for infinitesimal transformation with generator $G(x)$ one has to require $$ G(x)|\Psi (A_{\mu})\rangle = 0 $$ This reduces the Hilbert space by projecting it on a space with only physical gauge field polarizations. That's why the gauge symmetry is called a do-nothing transformation.
Next, suppose the "large" gauge transformation whose element $g_{(n)}$ carries a non-zero winding number $n$. We have that for the vacuum state on zero winding number configuration $|0\rangle$ $$ \tag 1 U(g_{(n)})|0\rangle = |n\rangle $$ One can introduce a $\theta$-vacuum defined as $$ |\theta\rangle =\sum_{n}e^{in\theta}|n\rangle, $$ so $$ \tag 2 U(g_{(n)})|\theta\rangle = e^{in\theta}|\theta\rangle $$ So we have that "large" gauge transformations are not do-nothing transformations; moreover, even after introducing the new vacuum it still acts non-trivially!
My questions are:
$(0)$ corresponds to invariance of classical gauge theory action under local gauge transformations. To which corresponds $(1)$? Naively I think that large gauge transformations change the gauge field strength tensor, but I would like to formalize this.
Finally, whether a classical analog of $(2)$ exists? Is this correspondence completely determined by classical gauge fields topology?