A moving charge has magnetic field around it. Is it just increased electric field in relativistic frame?
If yes, does a moving charge has more electric field around it?
Update: One shall take conservation of energy in account. Electric and magnetic fields have their own stored energy. Where does magnetic field energy come from in moving frame?
No, the magnetic field is in no way the "same thing" as an increased electric field. They are two (algebraically) independent fields $\vec E,\vec B$ at each point of space. The electric field attracts or repels a fox tail or an ebonite stick while the magnetic field attracts magnets out of iron. Those are different objects.
When we switch to another inertial system moving by the velocity $\vec v$, the transformation rules are most easily written when we split the electric and magnetic fields to parallel and longitudinal lines. $$ E'_\parallel = E', \quad B'_\parallel = B' $$ and $$ E'_\perp = \gamma (E_\perp -\vec v\times \vec B),\quad B'_\perp = \gamma (B_\perp +\frac{1}{c^2}\vec v\times \vec B) $$ where $\gamma = 1/\sqrt{1-v^2/c^2}$. If $\gamma\approx 1$ is a good approximation, we may just unify those to $$\vec E'=\vec E-\vec v\times \vec B,\quad \vec B' = \vec B + \frac{1}{c^2} \vec v\times \vec E$$ Note that the velocity creates new electric/magnetic fields that are of the opposite kind (electric vs magnetic) than the source from which they arose; and that are orthogonal both to the original field and the velocity.
If we substitute $\vec B =0 $ for the original magnetic field, we see that some new magnetic field is created; the longitudinal electric field along the velocity stays the same; and the transverse electric field is increased by the factor of $\gamma$.
I am not quite sure of what you ask. If you use Lorentz transformations, you can for example take the field of an electron moving at velocity $\vec{v}=-v\cdot\vec{e_{x}}$ in its rest frame, which is: $$\vec{E}(\vec{r_{'}})=e \cdot \frac{\vec{r_{'}}}{r_{'}^{3}}$$ Lorentz-transforming this into the lab frame yields: $$E(\vec{r})=\frac{e\gamma}{(y^2+z^2+\gamma^2 (x+vt)^2)^{ \frac{3}{2}}}\cdot (y \vec{e_{y}}+z \vec{e_{z}}+(x+vt) \vec{e_{x}})$$ $$B(\vec{r})=\frac{e\gamma\beta}{(y^2+z^2+\gamma^2 (x+vt)^2)^{ \frac{3}{2}}}\cdot(-y \vec{e_{z}}+z \vec{e_{y}})$$ So yes, a moving charge does have a magnetic field in a frame in which it is not at rest. This simply follows by the Lorentz transformations for the fields (which result from transforming the electromagnetic field tensor). So yes, magnetic and electric field depend on the frame, but this is an effect of the relativistic character of electrodynamics and there is no such thing as an increase in field as can be seen in the rest frame (even though of course, relativistically speaking we need to talk about the character of a field in a specific frame).
Update
I will try to adress the question in the comment. If there is a charge $q$ moving in positive x direction we need to follow the replacement rules: $$ v \to -v$$ $$ e \to q $$ In the frame in which the charge is moving, there is, like you said, a magnetic field given by above expression $B(\vec{r})$. There is also the electric field given by above $E(\vec{r})$. These are for example the fields we would measure in a laboratory after shooting out the charge with some experimental device. If we move along with the charge at the same velocity, we will not see any magnetic field, as the charge is stationary and by Maxwell's equations, stationary charges do not cause a magnetic field. We will however see a different electric field, $\vec{E}(\vec{r_{'}})$.
