Does $\nabla \phi = \nabla\times A$ have non-trivial solutions As stated in the title. Is there a non-trivial solution for $n=\nabla \phi$ and $n=\nabla \times A$. With the information that $|n|=1$ \begin{eqnarray} \nabla \times A = \nabla \phi \end{eqnarray} We can condense it to simpler expressions \begin{eqnarray} &\nabla^2 \phi =& \;0\\\ \nabla \times(\nabla \times A) =0\Rightarrow &\nabla^2 A=\;&\nabla (\nabla \cdot A)\\\ \nabla\times \nabla^2 A = \nabla^2(\nabla\times A)=&\nabla^2 n =&\;0 \end{eqnarray} So we have Laplace/Poisson equations. What would be the next step without digging into some representation?

Since the given conditions imply that $n$ is both divergence-free and curl-free, you are looking for a harmonic vector field; so in particular $\Delta n = \
abla \cdot \
abla n = 0$ as you deduced. The additional condition $|n|^2 = 1$ is now very powerful: the product rule gives $$\Delta |n|^2 = 2|\
abla n|^2 +2 n \cdot \Delta n = 2|\
abla n|^2,$$ so since we know $|n|^2$ is constant we conclude that $\
abla n = 0$ everywhere; i.e. $n$ is constant.