Does Noether's theorem apply to a strict on-shell symmetry of the action that holds on every integration region?

Question

I've worked through different proofs of Noether's theorem and read various posts about it on this site. Some important takeaways, among others from this and this post by Qmechanic were

1. Every off-shell$^1$ quasi-symmetry$^2$ of the action yields an on-shell continuity equation.
2. An on-shell quasi-symmetry is a tautology, i.e. every group of transformations is trivially an on-shell quasi-symmetry of the action.
3. It suffices to show that the (off-shell) quasi-symmetry holds for a single fixed integration region in order to apply Noether's theorem.

After studying this derivation it seems to me that if we have a strict symmetry$^3$ of the action for a specific field configuration which fullfills the EOM and if this symmetry holds on every region of spacetime, we get a continuity equation for this specific field configuration. My question is, if this statement is true.

I've read here that a strict symmetry of the action on-shell does not yield a conservation law. One possible interpretation of the latter statement is that it does not suffice to have a strict on-shell symmetry for a single fixed integration region in order for Noether's theorem to apply. This way, it wouldn't contradict my supposition. I'm not sure if this interpretation holds though, because in the same sentence Qmechanic claims that even a strict symmetry of the Lagrangian density on-shell does not yield a conservation law. However, according to my proposal, this should be the case. If the Lagrangian density has a strict on-shell symmetry then the action has a strict on-shell symmetry for every integration region.

Now I'm very confused because I haven't found the mistake in my arguments. As already explained, it seems clear to me from the referenced Wikipedia article that it is enough to consider a single field configuration which fullfills the EOM if we want to get a continuity equation for this field configuration. But the statements of Qmechanic seem to contradict this. So I would like to know if my proposition is correct and if not, why it does not work.

--

$^1$The words on-shell and off-shell refer to whether the Euler-Lagrange (EL) equations (=EOM) are satisfied or not.
$^2$A group of transformations that leaves the action invariant up to boundary terms.
$^3$A group of transformations that leaves the action invariant.

Answer 1

OP is right: In $n$ spacetime dimensions let there be given a first-order Lagrangian $n$-form $$\mathbb{L}~=~d^nx~{\cal L} .$$ If we have an on-shell strict$^1$ symmetry $$\delta x^{\mu}~=~\epsilon X^{\mu} \qquad\text{and}\qquad \delta\phi^{\alpha}~=~\epsilon Y^{\alpha}$$ of the Lagrangian $n$-form$^2$ $$ 0~\approx~\delta \mathbb{L}~=~\ldots~=~\epsilon d^nx~\left(\underbrace{E_{\alpha}}_{\approx 0 }Y_0^{{\alpha}} + d_{\mu}j^{\mu}\right)~\approx~\epsilon d^nx~d_{\mu}j^{\mu}, $$ where $j^{\mu}$ is the bare Noether current, then we also have an on-shell conservation law $$d_{\mu}j^{\mu}~\approx~0. $$

--

$^1$An on-shell quasi-symmetry is a triviality.

$^2$ The $\approx$ symbol means equality modulo the EL equations.