Riemannian metric adapted to singular $1$-dimensional foliation 
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.
 A: 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 $g$.
