What does the metric condition $\nabla_\rho g_{\mu\nu}=0$ in General Relativity intuitively mean for an observer measuring distances?

In General Relativity, the following condition hold: $\nabla_\rho g_{\mu\nu}=0$, where $g_{\mu\nu}$ is the metric of spacetime which has to do with measuring distances and angles and $\nabla$ is the covariant derivative which is torsion-free. What does this condition intuitively mean for an observer that moves on a manifold and measures distances on it?

I will give it a try but forgive me if I am a bit imprecise on some points as I try to see this intuitively and I feel that I don't get this quite right.
I start with $\partial_\rho g_{\mu\nu}=0$ (everywhere) which says that the metric is constant all over the manifold in question. Now, since the covariant derivative is an intrinsic derivative to the manifold, which roughly means that it is the change that a local observer on the manifold would experience, I would conclude that $\nabla_\rho g_{\mu\nu}=0$ means that an observer living on the manifold does not experience any change in the metric. I guess this seems pretty intuitive since an observer using a measuring tape to measure distances, won't experience any local change in the distances of objects around him/her.

Is this way of thinking about it right or is it imprecise? Any additional input would be appreciated.

Also, if an interpretation like the one above holds (or something similar), why do we need to have a torsion-less covariant derivative in order for the interpretation to have the meaning that it has?

The condition $\nabla_{a}g_{bc} = 0$ is just pure mathematics. Every metric admits a torsion-free (for one defintition, one that satisfies $\nabla_{[a}\nabla_{b]}f = 0$ for every function on the manifold) connection that satisfies this condition.
That general relativity is formulated using this connection is a statement that gravity obeys the equivalence principle -- a freely falling observer is parallel translated along the geodesics of $\nabla$ relative to $g$. And the fact that this is a parallel translation is encoded in that condition in the metric.
• So, can we say that an observer holding a rod and moving on a manifold, does so by parallel transporting himself and $\nabla_\rho g_{\mu\nu}=0$ means that s/he always sees constant angles and distances for the rod? If so, why, intuitively speaking, would a torsion-full covariant derivative break this physical picture? – TheQuantumMan Jul 31 '18 at 19:07