Does there exist a Schrödinger equation for the energy space, like for momentum? I would say no, because the energy basis is countable, but are there any other reasons?
Quantum Mechanics – Schrödinger Equation in Energy Basis
energyhilbert-spacequantum mechanicsschroedinger equation
Related Solutions
There are nonlinear versions of the Schrodinger equation that are completely irrelevant to your question. These are like the Gross-Pitaevski equation, they are nonlinear classical field equations that describe the flow of a self-interacting superfluid or BEC. These equations have nothing to do with the evolution of probability amplitudes, and I will not consider them further.
Probability theory is exactly linear
To understand why the concept of a nonlinear equation for probability amplitudes is not reasonable, and most likely completely impossible, consider first classical probability. Suppose I have a classical equation of motion of the form
$$ {dx\over dt} = V(x)$$
where the vector field V describes the future behavior as a flow on phase space, coordinatized by x. Now I can ask what is the evolution of a probability distribution $\rho(x)$, if I have incomplete knowledge of the initial position.
The evolution equation is determined by considering the probability of ending in a little box surrounding x'. This probability is the sum of all possible paths that lead to x' times the probability of being at the beginning of the path. This sum gives the probability equation:
$${\partial \rho\over \partial t} = V(x) \cdot {\partial \rho \over \partial x} - \rho(x)\nabla\cdot V $$
The point is that this equation is exactly linear, for fudamental reasons. It is impossible to even conceive of a nonlinear term in the evolution equation of a probability distribution, because the very definition of probability is lack of information, as represented by a linear space.
Note that classical probability distributions are defined on the entire phase space, so they are enormous dimensional linear equations which completely include the nonlinear dynamics if you restrict to delta-function sharp probability distributions on x. The only difference with quantum mechanics is that there are no delta-function sharp distributions in the presence of non-commuting observables on all observables. Otherwise the two types of descriptions are similar
Quantum mechanics mixes amplitudes and probabilities
If you have a quantum mechanical system, the wavefunction mixes with classical probability in a nontrivial way. If you consider a quantum system of two entangled spin 1/2 particles in a spin singlet, the projection of the wavefunction onto one of the two particles is a density matrix which is a classical probability.
This is extremely important to preserve, because the probabilities are nonlocally correlated, so if there were any way to extract the far-away component of the spin wavefunction, you would be almost certainly be able to use this to signal faster than light, because you can collapse the wavefunction where you are, and the far-away density matrix would then not have a probability interpretation.
These types of nonlinear theories are so difficult to conceive, that Weinberg suggested in the 1960s that quantum mechanics has absolutely no deformation of any kind which is consistent with no-signalling. Although this conjecture is not proved, to my knowledge, it is certainly plausible, and there are no nonlinear deformations which could serve as counterexamples (the link to this paper has just been posted as I write by Oda).
It is wrong to think that there is any nonlinear deformation of the Schrodinger amplitude equation. Such modifications do not exist, and almost certainly cannot exist. If the world obeyed such an equation with a tiny nonlinearity, different Everett branches would become interacting, and we would be able to see the ghosts of our other selves, and other nonsense. It would rule out any form of hidden-variable interpretation of the wavefunction, and it would almost certainly lead to violations of no-signalling.
As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation.
Recalling that
$$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$
and putting this expression into the (coordinate representation of the) TDSE, we have
$$i\hbar\frac{\partial}{\partial t}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) + V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$
We can move the partials inside the integrals but we must be careful with the potential. Noting that
$$\Phi(p,t) = \int dx \frac{e^{-i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \Psi(x,t)$$
we have
$$V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)$$
But,
$$ \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)= V(p) * \Phi(p,t)$$
where $*$ denotes convolution. Thus, we can write
\begin{align} \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left(i\hbar\frac{\partial}{\partial t}\Phi(p,t)\right) &= -\int dp \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)\right)\\&\qquad + \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left( V(p) * \Phi(p,t)\right) \end{align}
leading to
$$\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \left\{ i\hbar\frac{\partial}{\partial t}\Phi(p,t) = -\frac{\hbar^2}{2m}\left(-\frac{p^2}{\hbar^2}\Phi(p,t) \right) + V(p)*\Phi(p,t)\right\}$$
and so
$$i\hbar\frac{\partial}{\partial t}\Phi(p,t) = \frac{p^2}{2m}\Phi(p,t) + V(p)*\Phi(p,t)$$
Best Answer
The basis-independent form of the Schrödinger equation is $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\lvert\alpha\rangle = H\lvert\alpha\rangle.$$
If we wish to express this in a particular basis, we simply multiply by the corresponding basis bras. Thus, in the energy basis, we get \begin{align} \langle n\rvert i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\lvert\alpha\rangle &= \langle n\rvert H\lvert\alpha\rangle\\ i\hbar\dot c_\alpha(n) &= E_n c_\alpha(n), \end{align}
where $c_\alpha(n)$ is the energy-space wavefunction for the state $\lvert\alpha\rangle$, and $\lvert n\rangle$ are the energy eigenstates with eigenvalues $E_n$. Here, we can see that the energy basis is actually quite nice to work with, due to the fact that the time derivative of $c_\alpha(n)$ only depends $c_\alpha(n)$ and not on the value of $c_\alpha$ at other "points" (energy levels).