= Solution
First lower the index of the <four-current> with the <Minkowski metric>: $J_\beta=g_{\beta\alpha}J^\alpha$. The displayed expression is then
$$
\frac{\partial x^\beta}{\partial x'^\delta}J_\beta=J'_\delta.
$$
Therefore
$$
\boxed{\text{It is a covariant four-vector component, }J'_\delta.}
$$
Its one free lower index transforms with the inverse coordinate Jacobian. The contraction with the metric here lowers an index; it does not make the current a scalar because the index $\delta$ remains free.
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