Note that the finite transformation of:
$$
W^a_\mu \to W^a_\mu + \frac{1}{g} \partial_\mu \theta^a + \epsilon^{abc} \theta^b W^c_\mu
$$
is:
$$
W^a_\mu t^a \to g W_\mu^a t^a g^{-1} + \frac{i}{g} \partial_\mu g \tag{1}
$$
where:
$$
g = \exp(-i \theta^a t^a) \;\;\; \text{and} \;\;\; [t^a,t^b] = i \epsilon^{abc} t^c
$$
Thus, the first term on the right-hand side of equation $(1)$ transforms under the adjoint representation of the Lie group. The second term does not transform under the adjoint representation, but it should be easy to verify that the transformed gauge field still takes values in the Lie algebra (hint: looking at infinitesimal transformations is the easiest method to verify this).

In case you want more information on the adjoint representation of the Lie group, it might be worth looking at this question.

The first answer to such a question must always be: A gauge symmetry has no "physical" meaning, it is an artifact of our choice for the coordinates/fields with which we describe the system (cf. Gauge symmetry is not a symmetry?, What is the importance of vector potential not being unique?, *"Quantization of gauge systems"* by Henneaux and Teitelboim). Any gauge symmetry of the Lagrangian is equivalent to a *constraint* in the Hamiltonian formalism, i.e. a non-trivial relation among the coordinates and their canonical momenta.

In principle, any gauge symmetry may be eliminated by passing to the reduced phase space that has fewer canonical degrees of freedom. The gauge symmetry has *no physical meaning* in the sense that be may get rid of it by passing to a (classically) *equivalent description* of the system. A gauge transformation has *no physical meaning* because all states related by a gauge transformation are *physically the same state*. Formally, you have to quotient the gauge symmetry out of your space of states to get the actual space of states.

In contrast, a *global symmetry* is a "true" symmetry of the system. It does not reduce the degrees of freedom of the system, but "only" corresponds to conserved quantites (either through Noether's theorem in the Lagrangian formulation or through an almost trivial evolution equation in the Hamiltonian formalism). It is physical in the sense that states related by it may be considered "equivalent", but they are not the same.

Interestingly, for scalar QED, the global symmetry gives a rather inconvenient "Noether current" - one that depends on the gauge field (cf. this answer)! So the statement that "Noether's theorem" gives us charge/particle number conservation is *not* naively true in the scalar case (but it *is* in the Dirac case). Getting charge conservation from the gauge symmetry is also discussed in Classical EM : clear link between gauge symmetry and charge conservation.

Why then use such a "stupid" description in the first place, you might ask. The answer is that, in practice, getting rid of the superfluous degrees of freedom is more trouble than it's worth. It might break manifest invariance under other symmetries (most notably Lorentz invariance), and there can be obstructions (e.g. Gribov obstructions) to consistently fix a gauge. Quantization of gauge theories is much better understood in the BRST formalism where gauge symmetry is preserved and implemented in the quantum theory than in the Dirac formalism that requires you to be able to actually solve the constraints in the Hamiltonian formalism.

So the *key difference* between a gauge and a global symmetry is that one is in our *theoretical description*, while the other is a *property of the system*. No amount of shenanigans will make a point charge less spherically symmetric (global rotation symmetry). But e.g. the electromagnetic gauge symmetry simply vanishes if we consider electric and magnetic fields instead of the four-potential. However, in that case we *lose* the ability to write down the covariant Lagrangian formulation of electromagnetism - the current $J^\mu$ *must* couple to some other four-vector, and that four-vector is simply the potential $A^\mu$.

There is one further crucial aspect of gauge symmetries: *Every* massless vector boson *necessarily* is associated to a gauge symmetry (for a proof, see Weinberg's *"Quantum Theory of Fields"*). There is no other way in a consistent quantum field theory: You want massless vector bosons like photons - you get a gauge symmetry. No matter how "unphysical" this symmetry is - in the covariant framework of quantum field theory we simply have no other choice than to phrase such particle content in terms of a gauge field. This you might see as the true "physical" meaning of gauge symmetries from the viewpoint of quantum field theory. Going one step further, it is the *spontaneous breaking* of such symmetries that creates massive vector bosons. A theory of vector bosons is almost inevitably a theory of gauge symmetries.

As an aside: In principle, one might try to make any non-anomalous global symmetry into a gauge symmetry (cf. When can a global symmetry be gauged?). The question is whether gauging it produces any new *physical* states, and whether these states fit to observations.

## Best Answer

An

internal symmetryonly involves transformations on the fields of a theory, and must act the same independent of the point in spacetime. For example, consider a Lagrangian,$$\mathcal{L} = \partial_\mu \psi^\star \partial^\mu \psi - V(|\psi|^2)$$

for some potential $V$, and complex field $\psi$. The theory has an internal symmetry, namely one which rotates the field, i.e.

$$\psi \to \psi'=e^{i\alpha}\psi$$

where infinitesimally we would have $\delta \psi = i\alpha \psi$. The corresponding conserved current is,

$$j^\mu = i(\partial^\mu \psi^\star)\psi - i\psi^\star(\partial^\mu \psi)$$

which after quantization adopts the interpretation of charge or particle number.

A gauge transformation, on the other hand, is one which is dependent on the point in spacetime wherein one operates, and it may act on spacetime itself, or the fields. An example is the $U(1)$ gauge symmetry of electrodynamics, described by,

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

which is invariant under $A_\mu \to A\mu + \partial_\mu \lambda(x)$, for any function $\lambda(x)$. (To see this clearly, note $F=dA$, and hence the change $A \to A + d\lambda$ has no effect as $d^2\lambda=0$.)