what happens to a magnetic field when it moves relative to an observer at relativistic velocities? what happens to a magnetic field when it moves relative to an observer at relativistic velocities? Does it become an electrical field?
 A: Electric and magnetic fields are not invariant for observers in different frames of reference - and are simply components of a more fundamental "electromagnetic field".
In analogy to the Lorentz transformation of position and time in relativistic mechanics, there are also transformations that apply to electric and magnetic fields. So an observer in frame $S'$ moving with velocity ${\mathbf v}$ with respect to another frame $S$, will observe transformed fields given by (in SI units)
\begin{eqnarray}
{\bf E'} & = & {\bf E_{\parallel}} + \gamma \left[ {\bf E_{\perp}} + {\bf
v}\times {\bf B}\right]\, , \nonumber \\
{\bf B'} & = & {\bf B_{\parallel}} + \gamma \left[ {\bf B_{\perp}} -
\frac{1}{c^2} ({\bf v}\times {\bf E})\right] 
\end{eqnarray}
where the "parallel" and "perpendicular" subscripts refer to those components of the fields in the stationary frame that are directed parallel or perpendicularly to the relative motion between the frames of reference.
In the example in your question, you posit a situation where there is a magnetic field in the stationary frame (and I will assume, no E-field). In this case, we see from these transformations that in general you expect to see both a magnetic field and electric field in a moving frame of reference. 
In fact, it turns out to be impossible to "transform away" the magnetic field in this situation, because it turns out that $E^2 - c^2 B^2$ is a Lorentz-invariant. This invariant is negative in the stationary frame, but it could not be negative in any frame where the B-field is zero.
A: No, a purely magnetic field cannot become a pure electric field in another inertial reference frame. Here's a formula that relates the E and B fields from two inertial reference frames, S and S'.
Let's say that for an observer in S, there's only a magnetic field like in your case. Now for an observer in S' moving with a velocity $\vec v$ with respecto to S, the EM fields he will see are given by:
$\vec E'=\gamma (\vec E+\vec \beta \times \vec B) - \frac{\gamma ^2}{\gamma +1} \vec \beta (\vec \beta \cdot \vec E)$ and
$\vec B'=\gamma (\vec B-\vec \beta \times \vec E) - \frac{\gamma ^2}{\gamma +1} \vec \beta (\vec \beta \cdot \vec B)$ 
where $\vec \beta = \vec v /c$ and in your case $\vec E=\vec 0$ and so $\vec E' =\gamma  \vec \beta \times \vec B$, thus as long as the observer in S' isn't moving parallely to the $\vec B$ field, he will see an electric field.
But also $\vec B'=\gamma \vec B - \frac{\gamma ^2}{\gamma +1} \vec \beta (\vec \beta \cdot \vec B)$ and so the magnetic field according to the S' observer is different from the one for the S observer. 
Can $\vec B'$ be worth $\vec 0$? Well, no. Using the invariant quantity given by Rob Jeffries (setting $c=1$), we have that $|\vec E|^2-|\vec B|^2 = |\vec E'|^2-|\vec B'|^2 \Rightarrow -|\vec B|^2 = |\vec E'|^2-|\vec B'|^2$ and we see that we cannot set $\vec B'=0$ else a negative number ($-|\vec B |^2$) would be equal to a positive number ($|\vec E '|^2$) which is not possible.
