Action of 1-form symmetry

Question

1) Free scalar field

Let me start with simple illustration, how charge constructed from current and how charge acts:

$$ S = \int d^dx\; \partial_\mu \phi \partial^\mu \phi $$

We have obvious global symmetry: $\phi \to \phi + \alpha$. Conserved current from Nether procedure:

$$ J^\mu = \partial^\mu \phi $$

We can construct the charge by integration over space: $$ Q (t) = \int d^{d-1}x \; J^0 (t, x) $$

Using canonical commutation relations: $$ [\phi(t,x), \dot{\phi}(t,y)] = i\delta(x-y) $$

We immediately obtain: $$ [Q(t), \phi(t,x)] = -i $$ And for action on any operator: $$ [e^{i\alpha Q(t)}, \mathcal{O}(\phi(x))] = \mathcal{O}(\phi(x)+\alpha) $$

2) Interacting scalar field in 3d

Now let's turn to question.

$$ S = \int d^3x\; \left(\partial_\mu \phi \partial^\mu \phi + V(\phi)\right) $$

We have 1-form symmetry: $$ J_{\mu\nu} = \varepsilon_{\mu\nu\rho}\partial^\rho \phi $$

We can now integrate this current over 2d surface and simple obtain the same result as before, due to:

$$ J_{12} = \partial^0 \phi $$ And next all steps as in 1):

$$ Q = \int d^2x \,J_{12}(x) $$ $$ [e^{i\alpha Q(t)}, \mathcal{O}(\phi(x))] = \mathcal{O}(\phi(x)+\alpha) $$

But, action haven't such symmetry!!

1. I am only in starting point of understanding higher-form symmetry, so for me this looks contradictory.

Are my calculations and logic right?

2. Also in Generalized Global Symmetries exist statement that such operators acts on string-like objects. Could somebody clarify this?

This also related to One-form current in 3d QED .