What is the induced electromagnetic field of a point charge? If I move a point charge on some trajectory, then it will produce an electric field as well as a magnetic field. As the charge is moving, and as a point charge can not produce a steady current, then due to a varying current, an electromagnetic field will be induced. It will again produce another electric field, and also the magnetic field will induce electric field.
Now I am getting confused, how many electric field and magnetic are here. One electric field due to its own charge, on magnetic field due to its motion and another electric field due to the varying current.
Is the first electric field will also be occured even in magnetostatics? (due to motion of charge)
Do those two electric field follow principal of superposition? I mean do they superpose to form a ultimate resultant field?
 A: The field of a charge moving at constant speed is obtained by a Lorentz transformation from the Coulomb potential. For an accelerated point charge fundamental difficulties arise due to self acceleration. See https://en.m.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force.
A: When there are time varying fields we can write $E$ in terms of the vector potential $B=\nabla\times A.$ $$E=-\nabla V - \frac{\partial}{\partial t}A.$$
If we go to the Coulomb gauge, the term $E_{Coulomb}=-\nabla V$ is just due to the Coulomb potential of the charge as in ordinary electrostatics. And the second term $E_{induction}=-\partial A/\partial t$ is needed to satisfy Faraday's law so in that sense it is the field due to the changing magnetic field.
So yes the full $E$ field is the superposition of these two fields, but that is not a particularly useful way to think about it, because we still need to calculate $A$. It is not as easy to calculate $A$ as it is to calculate $V$ because the $B$ field depends on the changing $E$ field too (not just the current).
An easier way to figure out the $E$ and $B$ fields is to consider a point charge at rest, put the $E$ field ($B=0$ in this frame) into the electromagnetic tensor and boost with a Lorentz transformation to a moving frame.
A: 
In case of an electron most elementary it has an intrinsic static electric monopole charge $-e$ and an intrinsic static magnetic dipole charge or else called spin magnetic dipole moment approximately one Bohr Magneton in value ~$μ_{Β}$.
Note: The spin rotation illustrated above of the electron charge is an  intrinsic quantum spin and has no relation to the classical spin concept. The electron is according to the Standard Model a massive dimensionless point charge particle and cannot have a physical spin rotation. Charge of electron is depicted above as a yellow sphere just for illustration proposes and has pedagogical only value.
This electromagnetic charge of the electron creates the $E$ (electric Coulomb field) and $B$ (magnetic dipole field) static fundamental interaction fields of the electron with its environment.
The generated electric field total flux in space in V⋅m SI units is:
$$
\Phi_{E}=\unicode{x222F}_{S} \mathbf{E} \cdot \mathrm{d} \mathbf{S}=\frac{Q}{\varepsilon_{0}}
$$

*

*$\mathbf{E}$ is the electric field,

*$\mathbf{S}$ is any closed surface,

*$Q$ is the value of the electron elementary charge -e,

*$\varepsilon_{0}$ is the electric constant (a universal constant, also called the "permittivity of free space") $\left(\varepsilon_{0} \approx 8.854187817 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$
And magnetic flux on an open surface $S$:
$$
\Phi_{B}=\iint_{S} \mathbf{B} \cdot d \mathbf{S}
$$
The generated intrinsic spin magnetic dipole moment's total magnetic flux for the electron can be derived by this equation in Weber SI units:
$$\Phi_{μ_{Β}}=e c \mu_{0}$$
Where $e$ the absolute value of of the electron elementary charge, $c$ the speed of light and $\mu_{0}$ the vacuum permeability in SI units. The above equation gives the magnetic dipole flux for g-factor assumed to be 2 (i.e. anomalous magnetic dipole moment is neglected).
Also notice:
$$\frac{\Phi_{E}}{\Phi_{μ_{Β}}}=c$$
the speed of light.
And no, the electron does not need to translate (i.e. move linear in space) to generate a magnetic field. Both generated elementary static $E$ and $B$ fields of the electron's electromagnetic charge are intrinsic to the electron.
The statement that the electron charge has to translate in space (i.e. linear or orbital motion) in order to generate magnetism refers to the induced magnetism on non-magnetic electric current conductive matter like copper for example and not to the intrinsic static magnetism of the free electron which behaves like a tiny bar magnet even if hypothetically kept still and not moving.
In the case the electron is hypothetically not moving, kept still in space or moves at constant speed then there is no electromagnetic induction. Only when an electron accelerates we have EM radiation in the from of EM waves travelling in space that can cause EM induction on conductive matter.
Electrostatic and magnetostatic interaction of a free electron on air with its environment does not require from the electron to move. It can eletrostatically or magnetostatically interact with matter (i.e. depending also by the type of target matter if it can be electrically charged or magnetized)  even if the electron is hypothetically not moving.
Nevertheless, there is no still electron in nature.
