Is dynamics for each field necessary? Reading a paper by J. Bekenstein and A. Meisels on conformal invariance I found the following sentence, which I need to paraphrase to abstract the idea from its context.

Each field must be a dynamical field if the theory is to complete. One cannot hold that the field is prescribed in one gauge and is determined in others by means of its transformation law. The "original" gauge simply cannot be singled out from within the theory: all gauge choices are equally good before the laws Thus the theory must provide dynamics for each field

Is this really a necessity? If I write down a Lagrangian like
\begin{equation}
\mathcal L=\bar\psi i\gamma^\mu\partial_\mu\psi+eA_\mu \bar\psi\gamma^\mu\psi
\end{equation}
without providing the dynamics for $A_\mu$, I am not singling out a preferred gauge since the Lagrangian is gauge invariant. Experiments will then tell us the value of (gauge invariant combinations of) $A_\mu$ at any point.
If I had to make an objection to the above Lagrangian, it would be the lack of predictability.
P.S. The article is "Conformal invariance, microscopic physics and the nature of gravitation"
 A: I believe I see the mistake you're making here: the Lagrangian you write down is not, in fact, gauge invariant. This is a common misconception, but an important one to understand. I will first say a few things about this as the classical level, but I will also give a description of this issue as it appears in QFT since that's often where this issue first comes up, and based on the Lagrangian OP has chosen for an example, is something that OP has seen. Everything I plan to say will be directly in reference to OP's example Lagrangian, but none of the points I am looking to make are special to it.
Let's first think about what it means for $A_\mu$ to be non-dynamical. Of course, definitionally this means that no time derivatives of $A_\mu$ appear, so if we were to actually vary the action wrt $A_\mu$ to obtain its would-be equation of motion, we would get some nonsense which, in this case, would tell us that the field $\psi$ must just be zero (since the gamma matrices are non-degenerate). This is fine...there's nothing wrong with such a theory, but it's rather uninteresting, so this is almost always not the thing we want to do.
Instead then, we do not vary the action wrt the vector potential, opting to treat it as some known and given function instead. So if we were to take the variation wrt $\psi$, we would find its equation of motion, and that equation would depend upon this function $A_\mu$ which needs to be supplied in order to solve the equation. One thing we can say, however, is that, clearly, the solutions to the equation of motion will depend upon the $A_\mu$ that we choose. If we choose it to be zero, we are left with the free Dirac theory. If instead we choose it to be some other function, we will find solutions for $\psi$ which are no longer just plane waves...they will react to the "source" we have supplied.
Now let's think about my statement on the failure of this action to be gauge invariant. Writing the action as $S[\psi;A]=\int dx^4\mathcal{L}$ (I have used a semi-colon to remind us that $A$ is not on the same footing as $\psi$ since we need to supply it ourselves), we can ask ourselves what, very precisely, does gauge invariance actually mean? Normally, if we apply a gauge transformation with parameter $\lambda$, we would find (assuming $A_\mu$ is the vector potential for a $U(1)$ gauge group)
$$
\psi\rightarrow e^{iq\lambda}\psi,\ \ \ \ A_\mu \rightarrow A_\mu + \partial_\mu\lambda
$$
where I have denoted the charge of $\psi$ by $q$. The action we are thinking about here obeys the condition
$$
S[\psi;A] = S[e^{iq\lambda}\phi; A + d\lambda].
$$
That is, we need to change both $\psi$ and $A$ at the same time in order for the action to be invariant. If we only transformed the $\psi$ by itself, the of course this action would not come back to itself.
But in the context of everything I've said to this point, this should actually feel very strange...$A$ is a function we provided. Yet here we see that the action we are looking at only obeys gauge invariance if we manually intervene to change the function we are providing! It's not automatic, and that's an extremely important distinction. It also means that if we just went ahead and solved the equations of motion, to apply a gauge transformation we would not only need to change $\psi$ by a phase as normal, but also change the very function we supplied to find the solution...that means it's no longer a solution to the same equation we started with. The most important thing from all of this is, if you want something really concrete, is that this kind of "soft" invariance in which we need to manually reach in and intervene is not what Emmy Noether told us how to work with and so we lose all the power her theorems deliver to us.
So, that's all I think I want to say from the classical perspective. I think if we look at a path integral formulation, the evidence becomes even more damning. Define the path integral
$$
Z[A]=\int[\mathcal{D}\overline\psi][\mathcal{D}\psi]e^{iS[\psi;A]}.
$$
Of course, you may put any operators you like in the integrand to form correlation functions, none of that will be important to the point I would like to make. Note here that the partition function itself is a function of $A$ since we still need to specify it in order to know what the action we are working with is.
Now, since this is an integration, we can always change the dummy integration variable at no cost. In particular, there is nothing preventing us from changing the $\psi$ integrations to integrations over $e^{iq\lambda}\psi$ and the barred equivalent. These are just phase factors, so the Jacobian of the transformation won't concern us here. The important thing is that, under this change of variables, the action obeys
$$
S[e^{iq\lambda}\psi;A] = S[\psi; A - d\lambda].
$$
which follows from the invariance we discussed earlier.
But this fact implies that $Z$ is not invariant under a change of integration variable...rather it satisfies
$$
Z[A] = Z[A-d\lambda].
$$
So the partition function is very obviously not gauge invariant and hence no correlators you compute will be gauge invariant either. So there's really no good sense in which you could, at this point, claim the theory to be gauge invariant.
