Is the current in a loop accelerating in a uniform magnetic field the same for moving and stationary observer?

By Faraday's law, a changing magnetic flux through a loop produces electromotive force $\varepsilon=-\frac{\mathrm{d}\Phi}{\mathrm{d}t}$ . Suppose there is a rectangular loop in XY plane with Ohmic resistance $R$ and dimensions $a \times b$ moving at speed $v$ relative to an observer, in $x$ direction, in a uniform magnetic field $\vec{B}=b\vec{\hat{z}}$ . Magnetic flux through the a loop changes for the stationary observer due to length contraction $\Phi=B a (b / \gamma),$ inducing current $I=\varepsilon / R = \frac{B a b}{R \gamma^2} \frac{\mathrm{d}\gamma}{\mathrm{d}t}$ ( $t$ is time time in stationary frame). This means that when loop accelerates, it will have smaller area, so the flux changes, and therefore there is current induced in the loop. The formula is the same from the loop's perspective. If we gauge $\vec{B}$ as $\vec{B}=\nabla \times \vec{A}$ where $\vec{A}=B x \vec{\hat{y}}$ , we get that in loop's frame $A' = B (x' - vt') / \gamma \vec{\hat{y}}$ so $B'=B / \gamma$ , meaning $\Phi = (B / \gamma) a b$ and we recover the same result. Therefore current induced in the loop in both frames is the same $I=I'$ . But (four-)current should also transform under Lorentz transformations - what did I miss ? My first idea was that I am misunderstanding four current transformation and current is a scalar in this context. But if we now imagine the cross-section of the loop, and charge passing through that cross section in time $\delta t$ ( $\delta t'$ in moving frame) as $Q$ ( $Q'$ ). Then $\delta t' = \frac{1}{\gamma} \delta t$ by time dilation and $Q=Q'$ since it is scalar. So in that light it should be $I' = \gamma I$ so clearly I am missing something.