Can two different Schrödinger equations have the same wavefunction? Is it possible for two Schrödinger equations describing different systems to have the same wavefunction? And if that is the case, why or why not? 
 A: The question is very broadly posed, so there's a bunch of different ways to interpret it, and each of them can give a different answer.


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*Can two different Hamiltonians share an eigenfunction?
Yes. The answer by ComptonScattering gives an example in finite dimensions; if you want something closer to the usual three-dimensional quantum mechanics of a massive particle, there's even more examples (because the Hilbert space is much bigger). 
For something concrete, you can try the hydrogenic Hamiltonian $\hat H_0$ and $$\hat H = \hat H_0 + f(\hat r) \, \hat L{}^2,$$ where $f(r)\geq0$ is a non-negative function of the radius. Here the ground state will be the same for both (as will all the $\ell=0$ states) but $\ell\neq 0$ states will differ.

*Can two different Hamiltonians share all their eigenfunctions?
Yes. For a simple example, take any Hamiltonian $\hat H_1 = \hat H$ that is not the identity operator, and compare it with $\hat H_2 = \hat H{}^2$. Then every eigenfunction of $\hat H_1$ will be an eigenfunction of $\hat H_2$ (though the converse of that is not necessarily true), but the eigenvalues will in general differ.

*Can two different Hamiltonians share all their eigenfunctions and the eigenvalues?
No. This is because you can express the Hamiltonian as a function  of the eigenvalues and eigenfunctions. In Dirac notation, that reads $$\hat H = \sum_n E_n |n⟩⟨n|.$$

*Can two different Hamiltonians share at least one solution of the time-dependent Schrödinger equation, for all times?
Yes. The eigenfunction examples of the first point above are a suitable example.

*Can two different Hamiltonians share an arbitrary wavefunction in their solutions of the Schrödinger equation, at least for a single time $t_0$?
Yes. Easy: set up some arbitrary wavefunction $\psi(t_0)$, and let it run under your two arbitrary Hamiltonians $\hat H_1$ and $\hat H_2$. They'll typically take $\psi(t_0)$ in arbitrary, different directions, but hey, the solutions matched at time $t_0$.

*Can two different Hamiltonians share solutions of the time-dependent Schrödinger equation for all times and for arbitrary initial conditions?
No. Take an arbitrary wavefunction $\psi_0$, and use that as the initial condition $\psi(t_0) = \psi_0$ for the Schrödinger equation under arbitrary $\hat H_1$ and $\hat H_2$. By assumption, both solutions are equal, and in particular their time derivatives at $t_0$ are also equal. (In fact, that is all you need: agreement of the wavefunction and its time derivative, for arbitrary initial conditions.) This then implies that $$\hat H_1 \psi_0 = i\hbar \partial_t \psi(t_0) = \hat H_2 \psi_0,$$ i.e. that $\hat H_1$ and $\hat H_2$ agree on your arbitrary initial condition $\psi_0$. This then means that they must agree as operators.
As a final note, please note the huge number of questions that folded into your original query, because of its imprecise phrasing, and use it to learn the importance of providing sharp, well-defined questions.
A: Interpreting the question as "Can two different time independent Schrodinger equations have the same solution", or "... same ground state solution". 
Then yes by construction. Let $| \psi_i \rangle$, $i = 1,2,3$ be three orthogonal wavefunctions.
We can construct two Hamiltonians
$$H_a = E_1 | \psi_1 \rangle \langle \psi_1 | + E_2 | \psi_2 \rangle \langle \psi_2 | + E_3 | \psi_3 \rangle \langle \psi_3 |$$
and 
$$H_b = E_1 | \psi_1 \rangle \langle \psi_1 | + \frac{E_2}{2} \left( | \psi_2 \rangle \langle \psi_2 | + | \psi_3 \rangle \langle \psi_3 | + | \psi_3 \rangle \langle \psi_2 | + | \psi_2 \rangle \langle \psi_3 |  \right) + \frac{E_3}{2} \left( | \psi_2 \rangle \langle \psi_2 | + | \psi_3 \rangle \langle \psi_3 | - | \psi_3 \rangle \langle \psi_2 | - | \psi_2 \rangle \langle \psi_3 |  \right)$$
have the same ground state for $E_1<E_2,E_3,0$ (and the same spectra).
