= Solution
The canonical momentum for the stated symmetric-gauge <Lagrangian> is
$$
\mathbf p=m\dot{\mathbf x}+\frac q2\mathbf B\times\mathbf x,
\qquad
\frac{\partial L}{\partial\mathbf x}=\frac q2\dot{\mathbf x}\times\mathbf B.
$$
Since $\mathbf B$ is uniform and time independent, its <Euler-Lagrange equation> becomes
$$
m\ddot{\mathbf x}+\frac q2\mathbf B\times\dot{\mathbf x}
-\frac q2\dot{\mathbf x}\times\mathbf B=0,
\qquad\boxed{m\ddot{\mathbf x}=q\dot{\mathbf x}\times\mathbf B.}
$$
This is the magnetic <Lorentz force>, which is unchanged by constant spatial translations. The <Lagrangian> itself changes by
$$
\delta L=\epsilon\frac q2\dot{\mathbf x}\cdot(\mathbf B\times\hat{\mathbf n})
=\epsilon\frac{dF_{\mathbf n}}{dt},\qquad
\boxed{F_{\mathbf n}=\frac q2\mathbf x\cdot(\mathbf B\times\hat{\mathbf n}).}
$$
A total time derivative changes the action only by $\epsilon[F_{\mathbf n}(t_f)-F_{\mathbf n}(t_i)]$. For variations with fixed endpoints this boundary contribution has zero variation, so the equations of motion are unchanged. The translation is therefore a symmetry of the dynamics even though this particular gauge makes the <Lagrangian> only quasi-invariant.
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