Throughout this post I will be using Weinberg's notation. When a global symmetry group $G$ is spontaneously broken down to a subgroup $H$, it is often useful to reparametrize whatever fields that make up your theory in terms of fields which better reflect the low energy degrees of freedom. This reparameterization is often referred to as the coset construction of the low energy effective action.
In the coset construction, there a few objects that can be constructed, which ensure that whatever action you write down with them will remain invariant under the full group $G$, even though it is only manifestly invariant under $H$. In the interest of making this question compact, I won't go through all of the justifications of the construction, but simply state the results that are relevant.
In the end, you will arrive at three objects: $H$-matter fields $\tilde{\psi}$, a notion of an $H$-covariant derivative $\mathcal{D}_{\mu}=\partial_{\mu}+it_iE_{i\mu}$, where $t_i$ are the generators of $H$, and $E_{i\mu}$ is built out of the goldstone fields, and $D_{a\mu}$ fields which encode the derivative of the goldstone fields in an $H$-covariant way. The theorem is that any $H$ invariant lagrangian built out of $\tilde{\psi}$, $\mathcal{D}_{\mu}\tilde{\psi}$, and $D_{a\mu}$ is invariant with respect to $G$.
In this construction, we arrive at $E_{i\mu}$ which looks very much like an $H$-gauge field, in that it transforms inhomogeneously with respect to $g\in G$ $$t_iE_{i\mu}\to h(\xi,g)\big(t_iE_{i\mu}\big)h^{-1}(\xi,g)+i(\partial_{\mu}h(\xi,g))h^{-1}(\xi,g)$$ Where $h(\xi,g)\in H$ depends on the Goldstone fields $\xi_a(x)$. For completeness, here are the other transformations under $g\in G$
$$\tilde{\psi}\to h(\xi,g)\tilde{\psi}$$ $$x_aD_{a\mu}\to h(\xi,g)\big(x_aD_{a\mu}\big)h^{-1}(\xi,g)$$
With $x_a$ the broken generators. Essentially, the global $G$ invariance "looks" like an $H$-gauge symmetry (but it is not).
Although the original symmetry (global $G$) is not a gauge symmetry, everything is constructed in a way so that if we instead transform all of the fields with an $H$-gauge transformation
$$t_iE_{i\mu}\to h(x)\big(t_iE_{i\mu}\big)h^{-1}(x)+i(\partial_{\mu}h(x))h^{-1}(x)$$ $$\tilde{\psi}\to h(x)\tilde{\psi}$$ $$x_aD_{a\mu}\to h(x)\big(x_aD_{a\mu}\big)h^{-1}(x)$$
The action is invariant! My question is, is this a real symmetry? Are there real consequences of this? Or is this just some artifact that doesn't mean anything?