Is the vacuum of a local ${\rm U(1)}$ gauge theory unique?

Consider a spontaneously broken scalar field theory with a global $${\rm U(1)}$$ symmetry described by the Lagrangian $$\mathscr{L}=(\partial_\mu\phi^*)(\partial^\mu\phi)-\mathcal{V}(\phi),\\ \mathcal{V}(\phi)=\mu^2\phi^*\phi+\lambda(\phi^*\phi)^2$$ with $$\lambda,-\mu^2>0$$. The determination of classical vacua, in this case, amounts to the minimization of the potential. One finds that the vacua of this theory are infinitely degenerate.

Now consider a local $${\rm U(1)}$$ gauge theory described by $$\mathscr{L}=(D_\mu\phi^*)(D^\mu\phi)-\mathcal{V}(\phi)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ also degenerate or unique? Here, $$A_\mu$$ is the gauge field and $$D_\mu=\partial_\mu-ieA_\mu$$. This Lagrangian is invariant under $$\phi(x)\to e^{-ie\theta(x)}\phi(x),\\A_\mu\to A_\mu+\partial_\mu\theta(x).$$

Is the vacuum of the local theory described above unique both classically and quantum mechanically? If so, what is the best and/or simplest way to understand it?

I must also say why am I interested in this question. Elitzur's theorem points out that local symmetries cannot be spontaneously broken. Also we know that the necessary requirement of spontaneous symmetry breaking is the existence of a degenerate vacua. So I wonder whether local theories have unique vacuum (or is it that despite the vacuum being degenerate, local symmetries cannot be broken.). @DominicElse answer here gave me an impression that the the vacuum may be unique.

• @SRS The global U(1) transformations are not gauge transformations (since they do nothing to the gauge field - if you completely fix the gauge, there is still this global U(1)), the global U(1) here is precisely analogous to the global $\mathbb{Z}_2$ in Dominic Else's answer. – ACuriousMind Jun 29 at 16:16
• @SRS A valid vacuum configuration is not just $A=0$, but any $A$ that is pure gauge, i.e. $A=\mathrm{d}f$ for any function $f$, since all these $A$ have $F=0$. – ACuriousMind Oct 16 at 13:25