 Solution

For translation in a fixed direction $\hat{\mathbf{n}}$, take $\mathbf{K}=\hat{\mathbf{n}}$ and the preceding $F_{\mathbf{n}}$. The triple-product identity gives $F_{\mathbf{n}}=-(q/2)\hat{\mathbf{n}}\cdot (\mathbf{B}\times \mathbf{x})$. ThusChoosing the three Cartesian directions gives the three constantsThese are magnetic pseudomomentum components, not the three components of kinetic momentum $m\dot{\mathbf{x}}$. A direct check is $d{\mathbf{K}}_{\mathrm{mag}}/dt=q\dot{\mathbf{x}}\times \mathbf{B}+q\mathbf{B}\times \dot{\mathbf{x}}=0$. In the stated gauge, ${\mathbf{K}}_{\mathrm{mag}}=\mathbf{p}+(q/2)\mathbf{B}\times \mathbf{x}$, consistent with the Noether boundary correction.