Let $X$ be the following vector field on $\mathbb{R}^2\setminus \{0\}$
\begin{align}
x' &= x\,(1-x^2-y^2)(x^2+y^2-3) - y\,(2-x^2-y^2)\\
y' &= y\,(1-x^2-y^2)(x^2+y^2-3) + x\,(2-x^2-y^2).
\end{align}


[It is well known that](https://mathoverflow.net/questions/273970/a-cubic-system-with-two-nested-limit-cycles-with-opposite-orientations/274981#274981) there is no a Riemannian metric on the punctured plane such that the orbits of the above vector field would be unparametrized geodesics. [The proof is based on existence of opposite orientation of two consecutive nested limit cycles in the phase portrait of the vector field](https://mathoverflow.net/questions/273635/finding-a-1-form-adapted-to-a-smooth-flow/273689#273689).


Now we reduce the above geodesibility requirement to the following question;(Is this an obvious reduced question)?



> 
> Is there a (torsion free or metric or arbitrary) connection $\nabla$ on the punctured plane with $\nabla\_X X=0$?
> 
> 
> 



Is there a connection $\nabla$ for which this particular non geodesible vector field $X$ satisfy $\nabla_X X=0$?
Your vector field can be written in polar coordinates as 
$$X=X^r\partial\_r+X^{\varphi}\partial\_{\varphi}=r(4r^2-r^4-3)\partial\_r+(2-r^2)\partial\_{\varphi},$$
which exhibits the rotational symmetry.
The condition $\nabla\_X X=0$ implies that
\begin{align}
X^r\partial\_rX^r+\Gamma^r\_{rr}(X^r)^2+(\Gamma^r\_{r\varphi}+\Gamma^r\_{\varphi r}) X^rX^{\varphi}+\Gamma^r\_{\varphi\varphi}(X^{\varphi})^2=0, \\
X^r\partial\_rX^{\varphi}+\Gamma^{\varphi}\_{rr}(X^r)^2+(\Gamma^{\varphi}\_{r\varphi}+\Gamma^{\varphi}\_{\varphi r}) X^rX^{\varphi}+\Gamma^{\varphi}\_{\varphi\varphi}(X^{\varphi})^2=0.
\end{align}
For a connection with vanishing torsion, we have $\Gamma^r\_{r\varphi}=\Gamma^r\_{\varphi r}$ and $\Gamma^{\varphi}\_{r\varphi}=\Gamma^{\varphi}\_{\varphi r}$, which leads to the simplification
\begin{align}
X^r\partial\_rX^r+\Gamma^r\_{rr}(X^r)^2+2\Gamma^r\_{r\varphi} X^rX^{\varphi}+\Gamma^r\_{\varphi\varphi}(X^{\varphi})^2=0, \\
X^r\partial\_rX^{\varphi}+\Gamma^{\varphi}\_{rr}(X^r)^2+2\Gamma^{\varphi}\_{r\varphi} X^rX^{\varphi}+\Gamma^{\varphi}\_{\varphi\varphi}(X^{\varphi})^2=0.
\end{align}
Since the vector field has rotational symmetry, I will look for a connection with coefficients that depend only on $r$.
By making a fourth-order polynomial-in-$r$ ansatz for all Christoffel symbols,
$$\Gamma^i\_{jk}=\sum\_{m=0}^4\Gamma^i\_{jkm}r^m,$$
one can find the following solution:
$$\Gamma^r\_{rr}=-\frac{r}{2}, \quad \Gamma^r\_{r\varphi}=\Gamma^r\_{\varphi r}=\frac{1}{4}(3-12r^2+r^4), \quad \Gamma^r\_{\varphi\varphi}=0,$$
$$\Gamma^{\varphi}\_{rr}=2, \quad \Gamma^{\varphi}\_{r\varphi}=\Gamma^{\varphi}\_{\varphi r}=r(2-r^2), \quad \Gamma^{\varphi}\_{\varphi\varphi}=0.$$


It's worth noting that anything lower than a fourth-order polynomial will not produce any solution.
There does not seem to be any other connection with polynomial coefficients.


