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Usually when we say that the vectors are parallel, we say that it means that there's an equation ##k = \alpha l##, for the vectors ##k## and ##l## and some scalar ##\alpha##. In the context of differential geometry, the notion of a scalar means a scalar function of coordinates.

But if we take the geodesic equation ##u^\mu\nabla_\mu u^\nu = 0##, then if we consider a parallel vector ##\alpha u##, we have

$$\alpha u^\mu\nabla_\mu(\alpha u^\nu) = \alpha (u^\mu\partial_\mu\alpha)u^\nu + \alpha^2 u^\mu\nabla_\mu u^\nu = 0$$

So in general this is not equal to zero, although we can argue that it's proportional to ##u^\mu##, we have:

$$u^\mu\nabla_\mu u^\nu = f(x)u^\nu$$

But the geodesic equation in the general form ##u^\mu\nabla_\mu u^\nu = k(\lambda)u^\nu## is only preserved under affine transformations of the parameter, and so this new equation isn't the geodesic equation? So that means that congruences parallel to geodesics are not geodesics themselves? I kinda feel that this is trivial, but I had some conception in my head that this shouldn't be the case, that parallel vectors to the geodesic vector should also be geodesic, so I figured it's a misconception, but I wanted to check if I missed something in this conclusion.