= Solution
Yes. For any sufficiently smooth scalar $\chi$, the <gauge transformation>
$$
\boxed{\widetilde{\mathbf G}=\mathbf G+\boldsymbol\nabla\chi}
$$
leaves its <curl> unchanged because $\boldsymbol\nabla\times\boldsymbol\nabla\chi=\mathbf0$. This freedom is much larger than adding a constant vector: for example, $\chi=xy$ adds the nonconstant vector $(y,x,0)$. Two <vector potentials> for the same field differ by a curl-free field; on a <simply connected> domain that difference is a <gradient>, while nontrivial topology may allow additional global freedom.
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