What does the metric condition $nabla_rho g_{munu}=0$ in General Relativity intuitively mean for an observer measuring distances?

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.

I believe the best way to understand this is by the equivalence principle using a ruler and a clock. The equivalence principle says that within certain limits of distance and time, there is no way to determine whether or not you are freely falling toward a massive body, or floating in deep space far removed from all massive celestial bodies.

If the metric weren't a covariant constant, then your ruler and/or your clock would change their measurements. So, for example, you might move past an identically manufactured ruler lying perpendicular to the direction of travel, and your ruler would have a different length when you lay it parallel to the other one as you pass.

If you traveled around some closed path, your clock and ruler might not measure the same as when you left.

One of the arguments I encountered was that the scalar product should be invariant. But that just begs the question, why?

It says that the metric behaves as a constant with respect to covariant differentiation, meaning that vectors have constant magnitude under parallel displacement. In other words, if you displace a metre rule, it is still a metre rule, and if you displace a clock which measures one second per second (without disturbing the mechanism), it will still measure one second per second.

First of all, note that the covariant derivative of a vector is a tensor. This means that if we have two vectors $A_mu$ and $A^nu$, then their covariant derivatives $nabla_rho A_mu$ and $nabla_rho A^nu$ are tensors. These two tensors satisfy the transformation law: $$(nabla_rho A_mu)=g_{munu}(nabla_rho A^nu)$$ The same transformation can be applied to the two vectors $A_mu$ and $A^nu$ as follows: $$A_mu=g_{munu}A^nu$$ and take a covariant derivative of the transformation to obtain $$nabla_rho A_mu=(nabla_rho g_{munu})A^nu+g_{munu}(nabla_rho A^nu)$$ Now compare this with the first equation to observe that $$nabla_rho g_{munu}=0$$

Intuitive Meaning:

The metric tensor is thus conserved during covariant differentiation. However, this doesn't mean that the partial derivatives are all zero. One can express the partial derivatives of $g_{munu}$ in terms of the Christoffel connections and can make it zero only at a local point in the space-time. Using this process, one can show that the Christoffel connections are functions of the first derivatives of metric tensor.

One can obtain the expression for the Riemann curvature tensor which is a function of the first derivatives of the Christoffel connections (i.e., a function of the second derivatives of the metric tensor) to find that the Riemann tensor cannot be reduced to zero a local point. This means that all the second derivatives of the metric tensor cannot be reduced to zero at any local point of the space-time. The number of surviving second derivatives is $frac{1}{12}D^2(D^2-1)$, which for $D=4$ dimensions is 20. This is the number of independent components of the Riemann tensor.

The condition $nabla_rho g_{munu}=0$ intuitively means that all the first derivatives of the metric tensor can be zero at a local point in the space-time, but all the second derivatives cannot be identically zero at such a point. And the number of surviving second derivatives of the metric tensor determines the curvature of space-time. This also defines the very notion of locally-inertial frames.