Are possible gauge fields in a Lagrangian theory always determined by the structure of the charged degrees of freedom? An elementary example to explain what I mean. Consider introducing a classical point particle with a  Lagrangian $L(\mathbf{q} ,\dot{\mathbf{q}}, t)$. The most general gauge transformation is $L \mapsto L + \frac{d}{dt} \Lambda(\mathbf{q},t)$ which implies the usual transformations of the canonical momentum $p \to p+ \nabla_q \Lambda$. Generalizing this derivative as an extended one gives the connection of electromagnetism. Once the particle motion is quantized, we recognize this as a local $U(1)$ "internal" symmetry of the quantum-mechanical phase.
Is this a fundamental property of gauges symmetries - being implied by non-dynamical symmetries of the action? By "non-dynamical symmetries" I mean those coming from the structure of and the freedom of labeling for the degrees of freedom under consideration.  
EDIT: After reflecting on the comments below, I'd re-formulate the question as:  

Do non-dynamical symmetries of a local Hamiltonian exhaust all
  possible types of gauges fields that it can couple to in a
  gauge-invariant manner?

EDIT-2: The reason I'm asking is that it appears that the very possibility for a particle to couple to electromagnetism and gravity come from the applicability of the action formalism and is already built-in as the symmetry under addition of a total time derivative (which as I understand to be one of the possible general definitions of gauge symmetry).
Some comments suggest the answer is a trivial yes, presumably because non-dynamical symmetries are gauge symmetries by definition. A concise expert answer would be helpful to close the question.
 A: I have been trying to understand what you may possibly mean by a "non-dynamical symmetry" (which is surely not a term that is normally used in papers from "mainstream" authors, to put it politely) and I became convinced that it cannot mean anything. 
The problem arises in the third sentence when you write that the "most general gauge transformation" is
$$ L \to L+\frac{d\Lambda}{dt}.$$
But this is not a "transformation" in any sensible sense I can think of. This is a result telling you how the Lagrangian transforms under something – it transforms into itself up to a total derivative. But to define a transformation, you actually have to say how the fundamental fields $q,p$ actually transform, and not just how the Lagrangian transforms.
If a Lagrangian transforms to itself up to a total derivative, it means that the action
$$ S = \int dt\,L$$
may remain invariant given some favorable initial conditions at $\pm\infty$. So quite in general, it is allowed if symmetries transform the Lagrangian (or the Lagrangian density) up to itself plus a total derivative – or up to the divergence $\partial_\mu V^\mu$ in the field theory (multi-dimensional) case. In the component formalism (not superspace), this addition of total derivatives/divergences is inevitable e.g. for supersymmetry transformations.
But this result, how the Lagrangian itself transforms, is an extremely small part of the information that you need to actually define a transformation or a symmetry. So I don't think that you have defined any symmetry by saying how the Lagrangian transforms under it. There are infinitely many transformations that have this property.
The possibility to add a total derivative to the Lagrangian is completely general but specific gauge symmetries – such as Yang-Mills symmetry, diffeomorphisms, or local SUSY – are much more particular.
I think that the reason why you think that you're "deriving" a U(1) symmetry from the total derivative boils down to your confusing symbol $\Lambda$ whose total derivative is added to the Lagrangian. But the thing $\Theta$ whose derivative is added to the Lagrangian is a priori not the same thing as the parameter of a U(1) transformation. Instead, $\Theta$ may be an arbitrary complicated function of the fields (degrees of freedom) as well as the parameters of all the gauge transformations and perhaps derivatives of everything. 
For a simple collection of classical particles and a U(1) electromagnetic symmetry, $\Theta$ may be a simple function of $\Lambda$ only (it's actually the sum of $\Lambda(\vec x_i)$ evaluated at the positions of all the particles, and summed over these particles, so the relationship is not as trivial as you suggest); for other symmetries, it's a more complicated function. But you actually need to study how the degrees of freedom transform under a would-be gauge symmetry to determine whether it's there or not; you can't just look at how the Lagrangian should transform. When you do so, you discover Yang-Mills symmetries, diffeomorphisms, local SUSY, and a few others as sensible local symmetries. But this work can't be done just by looking at total derivatives.
