Riemannian metric adapted to singular $1$-dimensional foliation

Question

Is there a real analytic vector field $X$, locally defined around $0\in \mathbb{R}^{2n}$, with the following properties:

1) The origin is an isolated singularity for $X$ and its linear part is the matrix $J=\begin{pmatrix}0&I\\-I&0 \end{pmatrix}$.

2) There is a Riemannian metric locally defined on a neighborhood $U$ of the origin, such that all trajectories of $X$ in $U\setminus \{0\}$ are unparametrized geodesic.

This question is somehow a singular version of the concept "geodesible flow", a concept which is mainly related to Finding a 1-form adapted to a smooth flow, such that the flow has a unique isolated singularity whose linear part is in the form $\begin{pmatrix}0&I\\-I&0 \end{pmatrix}$.

Motivation: There is no such metric for $n=1$.

Edit: According to the comment of Jaap Eldering, I realized that the previous version of my question was elementary, so I've reformulated my question. I thank him for his comment.

Answer 1

No, this is not possible. In fact a more general result holds: If a vector field $X$ has an isolated singularity at $x\in M$ for which the linearization $X'(x):T_xM\to T_xM$ has no real eigenvalues, then there is no smooth Riemannian metric on an open neighborhood of $x$ for which all the integral curves of $X$ are unparametrized geodesics.

To see why, note that the integral curves of $X$ are unparametrized geodesics of a metric $g$ if and only if $\nabla^g_X X$ is a multiple of $X$, where $\nabla^g$ is the Levi-Civita connection of $g$. Meanwhile, if $x\in M$ is a zero of $X$, then one has $(\nabla^g X)(x) = X'(x)$ for any smooth metric on a neighborhood of $X$. In particular, if $X'(x)$ has no real eigenvalues, then $(\nabla^g X)(x)$ has no real eigenvalues, which implies that there is an open $x$-neighborhood $V$ (which depends on the choice of $g$) such that $(\nabla^g X)_y:T_yM\to T_yM$ has no real eigenvalues for any $y\in V$. In particular, $\nabla^g_X X = (\nabla^g X)(X)$ cannot be a multiple of $X$ at any point of $V$ other than at a zero of $X$. Thus, if $x$ is an isolated zero of $X$, then none of the integral curves of $X$ in $V\setminus \{x\}$ can be (unparametrized) geodesics of $g$.