Let $(M,g)$ be a Riemannian manifold, endowed with the Levi-Civita connexion $\nabla$ induced by $g$. By the very definition of the Levi-Civita connexion $\nabla$, we indeed know that $\nabla g=0$, i.e., the (total) covariant derivative of the metric tensor vanishes. Now assume that $G$ is another metric tensor field on $M$, such that it satisfies $\nabla G=0$, that is, the (total) covariant derivative of this alternative metric tensor vanishes, as well.


I have realised that trivially $G=k g$, where $k\in\mathbb{R}^+$, is a solution for $\nabla G=0$. However, I am interested to know whether this is the most general case, or else if there are other alternative metric tensors $G$ with vanishing covariant derivatives (with respect to the Levi-Civita connextion induced by $g$), which are not positive multiples of the original metric tensor $g$.



The vanishing of covariant derivative of an alternative metric tensor
In an irreducible Riemannian manifold $(M,g)$, every symmetric tensor that satisfies $\nabla^{g}T=0$ must be of the form $T=kg$ for some constant $k$. See Theorem 10.3.2 in Chapter 10 of Peter Peterson's *Riemannian geometry, 3rd Edition*. It is shown there how to use this fact to conclude that irreducible symmetric spaces are in fact Einstein manifolds.


In a reducible manifold, this mustn't be the case: given $(M,g)$ simply connected and complete, a theorem of de Rham tells us that it can be written as a product $(M,g)=(M\_{1}\times...\times M\_{n},g\_{1}+...+g\_{n})$, where the sum of the metric tensors is orthogonal, and each $(M\_{i},g\_{i})$ is an irreducible space. Then, one can define $T=k^{i}g\_{i}$ for any real constants $k^{i}$, which may be different from one another. Thus $T$ is not a multiplication of $g$ by a factor. If $(M,g)$ is not simply-connected or not complete, then this simply becomes a local argument, where $M$ and $M\_{i}$ are replaced with local simply connected neighborhoods.


The example above in the comments is a private case of this: a locally-flat space can always be locally decomposed into an orthogonal Riemannian products of $\mathbb{R}$.


I suspect that the other direction is also true, namely that every $\nabla^{g}T=0$ must be locally of this form. But I have no reference for this nor verified this myself.


