Is electric current relative? Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?
 A: Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.
A: $\let\g=\gamma \let\lam=\lambda$
I'd like to expand G. Smith's answer by applying it to a metal wire.
Assume a straight wire along $z$-axis. If it carries a current it will
contain stationary positive charges and moving electrons. The wire is
neutral, i.e. charge densities of positive and of negative charges
balance each other: 
$$\lam_+ + \lam_- = 0.\tag1$$ 
There will be a non-null current carried by electrons $I_-\ne0$,
whereas $I_+=0\ $ ($I = I_+ + I_-$). Note that by $\lam$ here I mean linear charge density, by $I$ the usual electric current.
It's useful to bear in mind signs. Of course $\lam_+>0$, $\lam_-<0$.
As to currents we may choose. If $I>0$ is taken, i.e. $I_->0$, then
$v<0$ (in order to have a positive current electrons have to move in
the negative direction).
As far as Lorentz transformations along $z$ are concerned, trasformation laws of $\lam$, $I$ are the same as for $\rho$, $J_z$:
$$\eqalign{
    \lam'_\pm &= \g \left(\!\lam_\pm - {v \over c^2}\,I_\pm\!\right) \cr
    I'_\pm &= \g\,(I_\pm - v\,\lam_\pm).\cr} \tag2$$
If we transform to average electrons rest frame, where $I'_-=0$, we have
$I_- = v\,\lam_-$ and substituting into the first of (2)
$$\lam'_- = \g \left(\!\lam_-  - {v^2 \over c^2}\,\lam_-\!\right) =
  {\lam_- \over \g}.$$
This is nothing but Lorentz contraction. Charge is invariant, so that
charge density transforms inversely to length. As far as electrons are
concerned, length in the lab frame is contracted wrt length in their
rest frame and charge density increases passing from the latter to the
former: $\lam_-=\g\,\lam'_-$. The opposite happens to $\lam_+$:
$$\lam'_+ = \g\,\lam_+$$
since $I_+=0$ and positive charges are at rest in the lab frame.
Summarizing, in electrons rest frame we have
$$\lam' = \lam'_+ + \lam_- = \g\,\lam_+ + {\lam_- \over \g} = 
  \left(\!\g - {1 \over \g}\!\right) \lam_+ = 
  \g\,{v^2 \over c^2}\,\lam_+ > 0$$
(I've used (1)). In that frame the wire is positively charged. As to current
$$I' = I'_+ + I'_- = -v\,\g\,\lam_+ = v\,\g\,\lam_- = \g\,I_- = \g\,I.$$
Of course this is the second of (2) written for $\lam$, $I$ with $\lam=0$.
We see that in electrons rest frame an increased current is observed (btw this also implies an increased magnetic field).
A: Charge density $\rho$ and current density $\vec{J}$ form a Lorentz four-vector $(c\rho, \vec{J})$ that transforms under a Lorentz transformation just like $(ct, \vec{r})$ does.
For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $v\hat{\mathbf{z}}$ relative to the unprimed frame, then the transformation is
$$\begin{align}
c\rho^\prime &= \gamma\left(c\rho - \frac{v}{c}J_z\right)\\
J_x^\prime &= J_x \\
J_y^\prime &= J_y \\
J_z^\prime &= \gamma\left(J_z - \frac{v}{c}(c\rho)\right)
\end{align}$$
where $\gamma=1/\sqrt{1-v^2/c^2}$.
If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.
Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.
