Schrodinger’s equation was originally just the partial differential equation satisfied by the position-space wave function of a particle (or more general system) in non-relativistic quantum mechanics. The same name is sometimes also used for the equation [math]\frac{d}{dt}\Psi(t)=iH\Psi(t)[/math] satisfied by the state vector in any NRQM system regardless whether or not a position-space representation is being used (or is even available).

It is deterministic (in the sense of determining [math]\Psi(t)[/math] uniquely for all [math]t[/math] if given an initial condition [math]\Psi(0))[/math], so long as the Hamiltonian [math]H[/math] is self-adjoint (symmetry is NOT enough!).

The proof of this involves more analysis than I could fit into a Quora answer, but in the general case it follows from the fact that for any self-adjoint operator [math]H[/math] on a Hilbert Space, the equation [math]\frac{d}{dt}\Psi(t)=iH\Psi(t)[/math] is uniquely satisfied by [math]\Psi(t)=e^{iHt}\Psi(0)[/math] where the complex exponential of [math]H[/math] is defined in terms of its spectral resolution; and for the PDE special cases it might be done by various theorems involving greens functions or Fourier analysis and convergence properties of improper integrals.

It may be non-deterministic if [math]H[/math] has not been specified on a large enough domain to be essentially self-adjoint (as sometimes happens if boundary conditions are omitted from the specification of a problem in which the particle is confined somehow – either by an infinite potential or by living in a single cell of a crystal lattice for example). But such cases are normally just due to inadequate specification of the problem rather than to a real physical indeterminacy.

So I would say that in a properly defined quantum theory model the Schrodinger equation is indeed almost always deterministic.

[N.B. It wasn’t part of the actual question, but I should perhaps add that the reason this does not make quantum mechanics deterministic is because even complete knowledge of the quantum state of a system is not sufficient to predict the outcomes of all possible experimental measurements. For any a state which happens to produce a predictable value for one observable there will be other observables for which the outcome is uncertain.]