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.