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The first necessary condition is that $\mathrm{tr}(R) = R^1_1+R^2_2 = 0$, otherwise there would be no parallel volume form (and hence no parallel metric). (This is one first-order differential equation on $\nabla$.)

Finally then, $\nabla$ is a metric connection in $U$ if and only if the identity $$ \mathrm{d}H + \mathrm{tr}(\gamma)\,H - H\gamma - {}^t\gamma H = 0 $$ holds. (This works out to be two second-order differential equations on $\nabla$, since the trace of $H^{-1}\mathrm{d}H$ vanishes by construction.)

Thus, a sufficient condition in this case is comprised by a first-order differential equation on $\nabla$, a strict inequality on the first derivatives of $\nabla$ (which is an open condition), and then two second-order differential equations on $\nabla$. Note that this is reasonable because a torsion-free connection depends locallly on $6$ functions of two variables while the metrics depend on $3$ functions of two variables, and one should expect $3 = 6-3$ differential equations to be be needed in a local characterization.

The first necessary condition is that $\mathrm{tr}(R) = R^1_1+R^2_2 = 0$, otherwise there would be no parallel volume form (and hence no parallel metric). (This is one first-order differential equation on $\nabla$.)

Finally then, $\nabla$ is a metric connection in $U$ if and only if the identity $$ \mathrm{d}H + \mathrm{tr}(\gamma)\,H - H\gamma - {}^t\gamma H = 0 $$ holds. (This is only four second-order differential equations on $\nabla$, since, by construction, the trace of $H^{-1}\mathrm{d}H$ always vanishes. One can check that these four equations are independent.)

Thus, a sufficient condition in this case is comprised by a first-order differential equation on $\nabla$, a strict inequality on the first derivatives of $\nabla$ (which is an open condition), and then four second-order differential equations on $\nabla$.

Now, suppose the open condition that $\mathrm{det}(R) = r^2 > 0 $ for some positive function $r$ on $U$ (which, if $\gamma$ is to be a Riemannian connection, would follow from $R\not=0$). Once this condition is imposed, define a symmetric matrix $H$ of determinant $1$ by the rule $$ H = \frac1r\begin{pmatrix}R^2_1& -R^1_1\\ -R^1_1 & -R^1_2\end{pmatrix}. $$ Finally then, $\nabla$ is a metric connection in $U$ if and only if the identity $$ \mathrm{d}H = - \mathrm{tr}(\gamma)\,H + H\gamma + {}^t\gamma H $$ holds. (This works out to be two second-order differential equations on $\nabla$, since the trace of $H^{-1}\mathrm{d}H$ vanishes by construction.)

In fact, since $\mathrm{d}\bigl(\mathrm{tr}(\gamma)\bigr)=0$, and $U$ is supposed simply-connected, there exists a positive function $f$ on $U$ such that $\mathrm{tr}(\gamma) = \mathrm{d}(\log f)$, and then the symmetric matrix $G = fH$ satisfies $$ \mathrm{d}G = G\gamma + {}^t\gamma G, $$ and so the metric $g = \pm G_{ij}\,\mathrm{d}x^i\mathrm{d}x^j$ (choose the sign so that $g$ is positive definite) is parallel with respect to $\nabla$.

Now, suppose the open condition that $\mathrm{det}(R) = r^2 > 0 $ for some positive function $r$ on $U$ (which, if $\gamma$ is to be a Riemannian connection, would follow from $R\not=0$). Once this condition is imposed, define a symmetric matrix $H$ of determinant $1$ by the rule $$ H = \pm\frac{1}r\begin{pmatrix}R^2_1& -R^1_1\\ -R^1_1 & -R^1_2\end{pmatrix}, $$ with the sign chosen to make $H$ be positive definite. Then $H$ is the unique symmetric positive definite matrix of functions on $U$ that has determinant $1$ and satisfies the condition that $HR$ be skew-symmetric.

Finally then, $\nabla$ is a metric connection in $U$ if and only if the identity $$ \mathrm{d}H + \mathrm{tr}(\gamma)\,H - H\gamma - {}^t\gamma H = 0 $$ holds. (This works out to be two second-order differential equations on $\nabla$, since the trace of $H^{-1}\mathrm{d}H$ vanishes by construction.)

In fact, since $\mathrm{d}\bigl(\mathrm{tr}(\gamma)\bigr)=0$, and $U$ is supposed simply-connected, there exists a function $f$ on $U$ such that $\mathrm{tr}(\gamma) = \mathrm{d}f$, and then the symmetric matrix $G = \mathrm{e}^f H$ satisfies $$ \mathrm{d}G = G\gamma + {}^t\gamma G, $$ and so the metric $g = G_{ij}\,\mathrm{d}x^i\mathrm{d}x^j$ is parallel with respect to $\nabla$, and, up to a constant factor, it is the unique such metric.

Thus, a sufficient condition in this case is comprised by a first-order differential equation on $\nabla$, a strict inequality on the first derivatives of $\nabla$ (which is an open condition), and then two second-order differential equations on $\nabla$. Note that this is reasonable because a torsion-free connection depends locallly on $6$ functions of two variables while the metrics depend on $3$ functions of two variables, and one should expect $3 = 6-3$ differential equations to be be needed in a local characterization.

For example, consider the $2$-dimensional case: Let $\nabla$ be a torsion-free connection on a simply-connected domain $U$ in the $x^1x^2$-plane, with connection coefficients $\Gamma^{i}_{jk}=\Gamma^{i}_{kj}$, let $\gamma^i_j = \Gamma^{i}_{jk}\,\mathrm{d}x^k$ be the entries of the $2$-by-$2$ matrix $\gamma$, and write $\mathrm{d}\gamma+\gamma\wedge\gamma = R\,\mathrm{d}x^1\wedge\mathrm{d}x^2$, where $R= (R^i_j)$ is a matrix of functions on $U$.

The first necessary condition is that $\mathrm{tr}(R) = R^1_1+R^2_2 = 0$, otherwise there would be no parallel volume form (and hence no parallel metric). (This is one first-order differential equation on $\nabla$.)

Now, suppose the open condition that $\mathrm{det}(R) = r^2 > 0 $ for some positive function $r$ on $U$ (which, if $\gamma$ is to be a Riemannian connection, would follow from $R\not=0$). Once this condition is imposed, define a symmetric matrix $H$ of determinant $1$ by the rule $$ H = \frac1r\begin{pmatrix}R^2_1& -R^1_1\\ -R^1_1 & -R^1_2\end{pmatrix}. $$ Finally then, $\nabla$ is a metric connection in $U$ if and only if the identity $$ \mathrm{d}H = - \mathrm{tr}(\gamma)\,H + H\gamma + {}^t\gamma H $$ holds. (This works out to be two second-order differential equations on $\nabla$, since the trace of $H^{-1}\mathrm{d}H$ vanishes by construction.)

In fact, since $\mathrm{d}\bigl(\mathrm{tr}(\gamma)\bigr)=0$, and $U$ is supposed simply-connected, there exists a positive function $f$ on $U$ such that $\mathrm{tr}(\gamma) = \mathrm{d}(\log f)$, and then the symmetric matrix $G = fH$ satisfies $$ \mathrm{d}G = G\gamma + {}^t\gamma G, $$ and so the metric $g = \pm G_{ij}\,\mathrm{d}x^i\mathrm{d}x^j$ (choose the sign so that $g$ is positive definite) is parallel with respect to $\nabla$.