Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 20 / a / ii / Solution

Yes. For any sufficiently smooth scalar $\chi$, the gauge transformationleaves its curl unchanged because $\nabla \times \nabla \chi =0$. 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.