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In my understanding neither John nor Legendre ellipsoids are scaling-invariant. As I explained in my answer, there is no scaling-invariant canonical construction. Since the formula is evidently scaling-invariant (by scaling we multiply $s$ and $h$ by the same constant and do not change $n$), it can not be canonical. There should be some misunderstanding from my side, I should think a bit.
Dean, let us together consider the square with side 2 centered at the origin and the mapping $(x,y) \mapsto (x/\sqrt{2}, \sqrt{2} \cdot y)$. This is my initial mapping scaling such that its determinant is $1$. We have the following sequences of your data $[n,s,h]$ for the initial square and its image w.r.t. to this transformation: $[(1,0), 2,1], [(0,1), 2,1]$ for the square and $[(1,0), 2\sqrt{2},1/\sqrt{2}], [(0,1), 2/\sqrt{2},\sqrt{2}]$ for image. Plugging these, we have $q(v)= 4 v_1^2 + 4 v_2^2$ for the square and $q(v)= 16 v_1^2 + v_2^2$ for its image. One is not image of the other.
Dean, I edited my ``answer'' without seeing your comment. I will redo your calculations. But in the newer version of my answer there is an explanation why you construction can not exist at all. I also changes the transformation in my counterexample slightly. Could you please look at it?
Could you please check your equation for the CONSTRAINED TRAJECTORY. For me it implies that the acceleration is proportional to the velocity which means that the trajectories are reparameterized geodesics of the metric $g$ (the choice of the function $V$ affects the "time" on the trajectory but not the trajectory considered as an unparameterized curve).
@Robert: The case $L^2=0$ and $L^2= -id$ were solved in Kručkovič, G. I.; Solodovnikov, A. S. Constant symmetric tensors in Riemannian spaces. (Russian) Izv. Vysš. Učebn. Zaved. Matematika 1959 1959 no. 3 (10), 147–158. (I did not check the whole arguments, and actually did not look on the case $L^2= -id$ at all, but the part of the arguments I have checked was OK and the general method is also OK).
@Robert: I added artificially the additional assumption that the metric is a cone metric, that is it has the form $dt^2 + t^2 g$, where $g$ is a metric on a 4D manifold. The assumption may sound artificial in this problem, but it is natural in the problem I was going to apply the possible answer on the question I have asked. Under this assumption (and also assuming $L^2=0$ and $rank(L)=2$, I calculated the components of $(g,L)$ in a certain coordinate system and to my surprise the image of $L$ is spanned over two parallel vector fields. There is no algebraic reasons for them though.
Dear Robert, I did today the calculation in a simpler case and observed the same phenomenon: the existence of one self-adjoint covariantly constant tensor implies the existence of additional properties that are differential and not algebraic consequences of this tensor. The simpler case is: the manifold is a flat cone over a 4D manifold of signature (2,2) and the g-selfadjoint (1,1)-tensor L has the property $L^2=0$ and has rank 2. Then, by calculations that required to go up to the second derivative of the curvature, one can show that the metric admits two lightlike parallel vectors.
