Propagation Of Magnetic Fields

what about a magnetic field created when a solenoid is energized with direct current? ... There is no frequency involved with direct current.

There is no frequency involved with steady direct current, but you are describing a switched direct current. This is also known as a step function, which does have frequency content at all frequencies: https://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html

what about permanent magnets. How do they propagate?

As you move a permanent magnet the changing B field produces an E field by Faraday's law. From there you have standard propagation of an EM wave.

Associating the curl of a magnetic field with rate of change of electric field, and the curl of an electric field with the rate of change of a magnetic field, is extremely useful, but not necessary.

[Note: the time derivative of the electric field does not cause a magnetic field, it causes the curl of a magnetic field, which is a big difference.]

In the Jefimenko formulation of electromagnetism, the electric and magnetic fields at $(vec r, t)$ are entirely caused by charges, currents, and their time derivatives...all at positions $vec r'$ consistent with the retarded time:

$$ t_r = t - frac{|vec r-vec r'|}c$$

There is no need to consider the state of the electric field when computing the magnetic field (and vice versa).

With that view point, a magnetic field some distance from a dipole, looks like a dipole field provided:

$$ frac{|vec r-vec r'|} c < t-t_r $$

Moreover, if you do consider the standard formulation with Maxwell's equations, it's important to consider what part of the field is radiative, and what is near-field. The dipole (magnetic) field of a bar magnet, $vec m$:

$$ vec B(vec r) =frac{mu_0}{4pi} big[ frac{3hat r(hat rcdotvec m) -vec m}{r^3} big]$$

falls as $1/r^3$. The energy falls as $1/r^6$, and is not radiative.