In chapter 5 of the book "Einstein Manifolds", Arthur Besse discusses the possibility to find the metric $g$ when knowing the Ricci curvature tensor $Ric(g)$ ($=R_{ij}$).
But what do we know about the problem of finding the metric by knowing the mixed Ricci tensor $R^i{}_j$?
The two problems seem to me different, because in the absence of the metric, the two tensors are not equivalent. A first step may be to find all possible symmetric tensors $g_{ij}$ which make $g_{ij}R^j{}_k$ symmetric, but this is far from solving the problem, because then $Ric(g)$ is not necessarily equal to $g_{ij}R^j{}_k$. So the next step may be to take each of these possible tensors $g_{ij}R^j{}_k$ and see if they are equal to $Ric(g)$, by using the known results for the $R_{ij}$ problem. But this seems much more complicated than the $R_{ij}$ problem. Perhaps it should be a more direct way.