quantum mechanics - Gauge transformation of vector potential multiplies wavefunction by phase

Consider an electron in an electromagnetic field with scalar and vector potentials $\phi, \mathbf{A}$. Suppose for simplicity that $\mathbf{A}$ is time independent. Suppose also that we know the wavefunction $\psi$ of this electron. Then $\psi$ satisfies

$$i \psi_t = \Bigg[ \frac{1}{2m} \left(\hat{\mathbf{p}}- \frac{e \mathbf{A}}{c} \right)^2 + e \phi \Bigg] \psi = \hat{\mathcal{H}} \psi$$

The question concerns showing that if you perform a gauge transformation of the potentials:

$$\mathbf{A} \rightarrow \mathbf{A}'= \mathbf{A} + \nabla \Lambda$$ $$\phi \rightarrow \phi' = \phi - \frac{1}{c} \frac{\partial \Lambda}{\partial t} = \phi$$

for some scalar $\Lambda (t, \mathbf{x})$, the wavefunction transforms as

$$\psi \rightarrow \psi' = \mathrm{exp} \left( \frac{i e \Lambda}{\hbar c} \right) \psi$$ i.e. it is multiplied by a phase. It is easy to show that $\psi'$ satisfies the transformed Schroedinger equation: $$i \psi'_t = \hat{\mathcal{H}}' \psi'$$

However, I would like to know if there are other possible solutions to the above equation. If so, what are they? Or, is $\psi'$ the only solution?

I tried to find other solutions by supposing $\psi' = f \psi$, where $f$ is unknown, and then plugging this into the new Schroedinger equation. This gives a new differential equation for $f$. However, so far my attempts at solving this differential equation have failed.

There is probably another way (perhaps via path integrals?) of showing this that I am not aware of. Could you give me a clue, please?