I had already asked this question in MSE then I ask here at MO.
Assume that $A\in M_n(\mathbb{R})$ is a non singular matrix.
Is the flow of the linear vector field $X'=AX$ a geodesible flow on $\mathbb{R}^n \setminus \{0\}$? Namely, is there a Riemannian metric on $\mathbb{R}^n \setminus \{0\}$ such that the trajectories of the linear vector field are unparametrized geodesics?
Remark: For $n=2$ the answer is affirmative, as we explain below:
Fact: A linear vector field associated to a non singular $2 \times 2$ real matrix is a geodesible vector field on the punctured plane.
Proof:
Let $A$ be an invertible matrix. We denote by $X$ the linear vector field associated to $A$.
We consider two cases:
$A^2$ has no real eigenvalue.
$A^2$ has a real eigenvalue.
Case 1) In this case the linear vector field $Y$ associated to the matrix $A^{-1}$ is transverse to $X$ on the punctured plane and satisfies $[X,Y]=0$. This obviously implies that $X$ is a geodesible vector field.
Case 2) If $A^2$ has a real eigenvalue then $A$ is similar to one of the following matrices:
$$\begin{pmatrix} a&0\\ 0& b \end{pmatrix};\; \begin{pmatrix} a&\epsilon\\ 0& a \end{pmatrix} ;\;\begin{pmatrix} 0&b\\ -b& 0 \end{pmatrix}.$$ For the first matrix the closed $1$-form $\psi=ax\,dx+by\,dy$ satisfies $\psi(X)>0$. So $X$ is a geodesible vector field. For the second matrix the $1$-form $\psi=ax\,dx+ay\,dy$ satisfies $\psi(X)>0$. For the third matrix the vector field is geodesible because we have a foliation of the punctured plane by closed curves.
The reason of geodesibility of case 1 and the three matrices in case 2 is discussed in Finding a 1-form adapted to a smooth flow, which is essentially based on page 71 of "Geometry of foliation " by Philip Toender, Propsitions 6.7 and 6.8.
Please see also the related post Is every real matrix conjugate to a semi antisymmetric matrix?.