quantum mechanics - Is the sign in the Schrodinger equation physical?

I always have trouble remembering the sign in factors like $\exp(\pm ik\cdot x)$ (I'll use mostly minus signature here) that arise in field theory.

My mnemonic is to remember that the Schrodinger equation is $i\partial_t|\Psi\rangle=H|\Psi\rangle$ and so the wavefunction goes something like $|\Psi\rangle\sim \exp(-i\omega t)$ where $\omega$ is the energy of the state and then I just remember that this is the same sign assigned to incoming particles in Feynman diagrams. Similarly, this is the same sign that comes with the annihilation operator when we expand out a field in terms of creation and annihilation operators. The sign in $|\Psi\rangle\sim \exp(-i\omega t)$ is then just determined by the sign appearing in the Schrodinger equation above.

My question is whether the sign appearing in the Schrodinger equation is physical? Relatedly, are the signs appearing in Feynman diagrams or more generally the overall sign in the path integral physical?

It would seem to me that all calculations would remain the same if we switched signs everywhere since physical quantities just depend on complex inner products and this would appear to be insensitive to flipping the signs everywhere..