Does the occurrence, not just the measurements, of events also depend on the state of motion of the observer? If a rod is moving with some velocity, its length is measured differently by different observers, depending on their velocity with respect to the rod. But, to all observers in all frames, there is a rod to measure, right?
Consider this example. An observer at rest sees a charge moving with a uniform velocity v and another observer is moving with the charge at the same velocity.
The resting observer sees a moving charge and hence sees a magnetic field due to this motion. However, the observer moving with the charge at the same velocity sees the charge to be at rest. Hence, this observer should not observe/detect/measure a magnetic field. Is this truly the case? Does one observer see/detect a magnetic field and the other does not?
The rod example illustrating length contraction may not be as similar as I think it is. But it seems to be a good example to illustrate what is meant by the title of the question.
 A: 
But, to all observers in all frames, there is a rod to measure, right?

Correct, the events that make up the world 'line' of the rod are invariant (the same for all observers) while the spacetime coordinates assigned to these events are observer dependent.

Hence, this observer should not observe/detect/measure a magnetic
field. Is this truly the case?

To say that the measured magnetic field strength of the charge is zero in a frame in which it is at rest isn't conceptually identical to saying that the magnetic field does not exist in that frame. That is, the left hand side of the equation $\mathbf{B} = 0$ refers to something, the magnetic field, and not nothing.
But even if one doesn't accept this metaphysical point, there is another way to think about this question that essentially gets rid of the question itself. The fact of the matter is that the electric and magnetic fields are not actually separate though related vector fields in the context of relativity. If they were, they would necessarily transform as four-vectors and so there would be separate electric and magnetic four-vectors.
Instead, the four-vector associated with electromagnetism is the electromagnetic four-potential $\vec{A}$ which combines the electric scalar potential $\phi$ with the magnetic vector potential $\mathbf{A}$:
$$\vec{A} = (\phi/c, \mathbf{A})$$
The exterior derivative of the electromagnetic four-potential produces an antisymmetric rank-2 tensor field called the Faraday tensor $F^{\mu\nu}$ containing 6 independent components (expressed here in Cartesian coordinates):
$$  F^{\mu\nu} = \begin{bmatrix}
     0     & -E_x/c & -E_y/c & -E_z/c \\
     E_x/c &  0     & -B_z   &  B_y    \\
     E_y/c &  B_z   &  0     & -B_x   \\
     E_z/c & -B_y   &  B_x   &  0
  \end{bmatrix}$$
See that the components of the electric and magnetic vector fields are actually components (appropriately scaled) of a single higher rank object, the electromagnetic field tensor. So, in the specific case of the single charge at rest, the electric components of the EM field tensor are non-zero while the magnetic components are zero and there is no question about the existence of the EM field tensor field itself.
