0) In your 2nd paragraph, I presume you meant to write "a changing magnetic field can induce an electric field" (Faraday's law.)
1) Your first question concerns, I think, a static charge distribution plus a set of changing magnetic fields (produced, perhaps, by a set of dynamic currents somewhere off-stage). In this case,
- the static charges will produce a conservative (curl-less) electric field.
- the changing magnetic fields will induce a divergence-less (non-conservative) electric field
The total electric field will be the sum of these two. (They obviously cannot be equal, since one has a divergence but no curl, and the other has a curl but no divergence.)
2) While a current is by definition charges in motion, magnetostatics studies cases where the currents do not change with time (and net charge density = 0). In those cases the current density $\boldsymbol{j}$ is constant in time, the resultant fields are magnetic fields that are also unchanging in time, and no changing electric fields are induced. So they're not being neglected, they just don't exist in these cases.
3) (your first bullet) The displacement current (aka Maxwell correction term) is required to maintain the consistency of the equation. The classic demonstration is of a capacitor being charged via wires. Here the current is constant but the charge on the capacitor plates is increasing linearly with time.
Imagine trying to calculate the line integral of the magnetic field around a path surrounding the wire, without the displacement current term.
- If one constructs a surface, bounded by the path, that intersects the wire, you can apply Stoke's theorem and Ampere's law and get a non-zero result.
- However, a second surface, also bounded by the path, that passes between the plates of the capacitor, has no current crossing it and so gives a zero result for the line integral.
The displacement current term removes this inconsistency, since the charging capacitor has a changing electric field between its plates.
4) (your 2nd bullet)
- In the static charge case, $V$ is just the normal Coulomb potential.
- For the time-varying case, the potentials are not unique: various possibilities, all related by gauge transformations, give the same fields. For example, the Coulomb gauge (in which $ \boldsymbol{\nabla \cdot A} = 0$) gives the same formula for the potential as for the static case. Other gauge choices will give different results. I'm sure your text discusses the gauge conditions.
A system with no moving charges is consistent with there being only a static electric field and no magnetic field. However, it does not require there to be no time dependent phenomena. The general solution in this case consists of an electrostatic field, plus freely propagating electromagnetic waves.
You can see the consistency of the static fields by setting the time derivatives in your equations to zero. Then there is only a static divergence source for $\vec{E}$, meaning just an electrostatic field. However, it is possible to add the time-dependent fields of one or more propagating waves in top of that.
In general, any system of differential equations is going to have its solutions determined by the equations themselves, along with the boundary conditions. It's the (space and time) boundary conditions in this case that determine whether there are also freely propagating electromagnetic waves present.
Best Answer
A constant charge density does not imply a zero magnetic field. Even considering a set of isolated charges, suppose they were (mechanically) moved along a circular path. The charge density could remain the same but there would be a current flow. The curl of the magnetic field produced would be $\mu_0 \vec{J}$, where $\vec{J}$ is the current density.
If the charge density is static, all you can say (from the equation of charge continuity) is that the divergence $\nabla \cdot \vec{J} = 0$. This also does not imply time-independence. For instance, you could speed up the circular motion of the charges to get a larger current density and a larger magnetic field.
If you can have a time-dependent magnetic field then you will also get a time-dependent electric field. The time-independence of $\rho$ only tells you that the divergence of the electric field is time-independent.