The requirement that parallel transport preserve the length of vectors is equivalent to requiring that $ \nabla_Z ( g(X,Y) )$ vanish for all vector fields $Z$ and all vector fields $X,Y$ that are autoparallel along the integral curves of $Z$, i.e. $\nabla_Z X = \nabla_Z Y = 0$.
It is clear to me that this condition is equivalent to the condition $\nabla g (X,Y,Z) = 0$ provided we again require that $\nabla_Z X = \nabla_Z Y = 0$. However, we want to show of course that metric compatibility requires $\nabla g = 0$ so we need to extend to arbitrary $X,Y,Z$.
I figure we can do this by looking at the function $\nabla g(X,Y,Z)$ point by point. At a point $p$ on our manifold the function is determined by the action of the tensor $\nabla g(p)$ on the vectors $X(p),Y(p),Z(p)$. We can then replace $X$ and $Y$ with fields $X'$ and $Y'$ which are equal to $X$ and $Y$ at $p$ and are additionally autoparallel along the integral curve of $Z$ crossing $p$. We then find that $$\nabla g(p) (X(p),Y(p),Z(p)) = \nabla g(p) (X'(p),Y'(p),Z(p)) = \nabla g(X',Y',Z)(p) = 0$$ from this we can infer that $\nabla g = 0$, as desired. However, this step of replacing $X$ and $Y$ with $X'$ and $Y'$ seems convoluted. Am I missing some shorter step which allows us to extend the argument to general $X,Y,Z$?