There isn't a mechanism. You're trying to find a mechanism for how two abstract objects can exchange identities. Any mechanism involving abstractions must consist of abstractions. So,the only way to explain it is through mathematics.
Least abstract way to look at it
I feel that the least abstract way to explain it is to look at two stationary charges. They attract via the Coulomb (electrostatic) force. Now run perpenducular to the line joining their centers. Each charge creates a magnetic field as it is moving (moving charge can be thought of in certain cases as current). The magnetic field acts upon the other charge, creating a force. Meanwhile, the electric force has decreased (no longer electtoSTATIC). The net force is the same, but part of it is magnetically caused.
Relativistic way
Another way to look at it is to remember that EM fields are set up by EM radiation. An EM wave carries oscillating EM fields with it; see pictures here. A point charge radiates EM waves in all directions. The oscillating E field of one of these waves interferes with the E field from a nearby wave constructively, creating a nearly non-oscillating field, which decreases as distance squared (Comes from the fact that intensity of a point source $\propto 1/r^2$), giving us Coulomb's law. The oscillating magnetic fields destructively interfere, so we get no net magnetic field.
Now, if you start moving with respect to the charge, things get interesting. The relativistic doppler effect will act upon the EM waves, altering them (since the speed of light is the same in all frames, we can't apply relative velocities to it). The interferences won't work quite the same, and we'll get a bit of a magnetic field and mainly an electric field. Move faster, and the magnetic field intensity increases, E decreases. Accelerate, and you get complicated stuff. Note that infact em waves are radiated only by an accelerated charge. A sitting charge does not emit em waves. The waves emitted by an accelerated charge produce change in the fields. The easiest way to visualize this is by assuming that the em waves are radiated in all cases.
I think that explains it without too many abstractions..
However, the following is confusing:
In a reference frame that sees the protons moving, the same forces
result from the magnetic force created by the protons' movement.
I agree that it is confusing. In the instant that the two protons begin to move away from each other they are momentarily mutually at rest. At this instant, in a frame of reference in which the proton's are (momentarily) at rest, the force on each proton due to the other is purely electric in nature.
At this instant, in this frame, the coordinate acceleration of either proton is equal to its proper acceleration (the acceleration as measured by an 'accelerometer' on the proton)
However, in the lab frame, special relativity dictates that the coordinate acceleration of the protons away from each other $(\mathbf{a}'_\perp)$ is actually less than the proper acceleration away from each other $(\boldsymbol{\alpha}_\perp)$:
$$\mathbf{a}'_\perp = \frac{1}{\gamma^2}\boldsymbol{\alpha}_\perp$$
But we also have that
$$\mathbf{F}'_\perp = \gamma m_p \mathbf{a}'_\perp$$
$$\mathbf{F}_\perp = m_p \boldsymbol{\alpha}_\perp$$
Thus, it must be that
$$\mathbf{F}'_\perp = \frac{1}{\gamma}\mathbf{F}_\perp$$
That is, in the lab frame, there is less of a repulsive force on each proton.
Since charge is Lorentz invariant, the electric force of repulsion between the protons is unchanged from the (momentary) rest frame and so, if there were only an electric force, there would be an inconsistency.
However, there is also an attractive magnetic force component in the lab frame and so, the electromagnetic force of repulsion between the protons is less in the lab frame and is thus consistent with SR.
To be sure, let's check the calculation. The fields at the location of the 'upper' proton due to the 'lower' proton in the momentary rest frame transform to the lab frame as
$$\begin{align}
& \mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}\\
& \mathbf {{B}_{\parallel}}' = \mathbf {{B}_{\parallel}} = 0\\
& \mathbf {{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) = \gamma \mathbf {E}_{\bot} = \gamma \frac{e}{4\pi\epsilon_0d^2}\hat{\mathbf{z}}\\
& \mathbf {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right) = -\gamma \frac{1}{c^2} \mathbf{ v} \times \mathbf {E} = +\gamma \frac{v}{c^2}\frac{e}{4\pi\epsilon_0d^2}\hat{\mathbf{x}}
\end{align}$$
where the lab frame has velocity $\mathbf{v} = -v\hat{\mathbf{y}}$ in the momentary rest frame.
Since the protons have zero velocity in the momentary rest frame, the Lorentz force on the 'upper' proton is
$$\mathbf{F} = e\left(\mathbf{E} + \mathbf{0} \times \mathbf{B} \right) = \frac{e^2}{4\pi\epsilon_0d^2}\hat{\mathbf{z}}$$
In the lab frame, the velocity of the protons is $\mathbf{u} = v\hat{\mathbf{y}}$ and the Lorentz force is
$$\mathbf{F}' = e\left(\mathbf{E}' + \mathbf{u} \times \mathbf{B}' \right)\frac{e^2}{4\pi\epsilon_0z^2} \gamma\left(1 - \frac{v^2}{c^2} \right)\hat{\mathbf{z}} = \frac{1}{\gamma}\mathbf{F}$$
as desired.
Best Answer
But at the same time if you take a magnetic dipole (a magnet as we know it) and move it around you will all of sudden get an electric field.
It was a great step forward in the history of physics when these two observations were combined in one electromagnetic theory in Maxwell's equations..
Changing electric fields generate magnetic fields and changing magnetic fields generate electric fields.
The only difference between these two exists in the elementary quantum of the field. The electric field is a pole, the magnetic field is a dipole in nature, magnetic monopoles though acceptable by the theories, have not been found.
Electric dipoles exist in symmetry with the magnetic dipoles:
$\hspace{50px}$
.$$
\begin{array}{c} \textit{electric dipole field lines} \\ \hspace{250px} \end{array}
\hspace{50px}
\begin{array}{c} \textit{magnetic dipole field lines} \\ \hspace{250px} \end{array}
$$
$\hspace{50px}$
There is symmetry in electric and magnetic forces
(the next is number 2 in the question)
Historically magnetism was observed in ancient times in minerals coming from Magnesia, a region in Asia Minor. Hence the name. Nothing to do with obvious moving electric fields.
After Maxwell's equation and the discovery of the atomic nature of matter the small magnetic dipoles within the magnetic materials building up the permanent magnets were discovered.
No. See answer to 2. Changing magnetic fields create electric fields and vice versa. No net charges involved.
No. A magnetic field interacts to firs order with the magnetic dipole field of atoms. Some have strong ones some have none. A moving magnetic field will interact with the electric field it generates with the electrons in a current.
A magnet has zero electric charge usually, unless particularly charged by a battery or whatnot. It has a magnetic dipole which will interact with magnetic fields directly. See link above.
It is an observational fact, an experimental fact, on which classical electromagnetic theory is based, and the quantum one. Facts are to be accepted and the mathematics of the theories fitting the facts allow predictions and manipulations which in the case of electromagnetism are very accurate and successful, including this web page we are communicating with.