Doppler shift in De Broglie's wavelength

If I were to take a particle, moving at a non relativistic velocity, it will have a wavelength (according to De Broglie). However, If I were to change the referance frame of the motion, I know that the wavelength will change because the momentum changes. But will the change be synonymous with the shift of wavelength, if I did the same problem, just considering it wave and applying doppler's shift.

If so, why? And if not so, then also, why?

What would be the changes if I were to take a relativistic frame instead of a non relativistic one?

PS: Due be noted that I am a high school students, who just has a surface level understanding of lorentz transformations and relativity.

If $$\psi(x,t)$$ is a solution of the time-dependent Schroedinger equation in a frame at rest, then, using the chain rule for partial derivatives, you will find that from a frame frame moving at speed $$-U$$ the solution becomes $$\tilde \psi(x,t)= e^{imUx/\hbar -i\frac 12 mU^2t/\hbar}\psi(x-Ut,t).$$ In particular, wavefunctions do not transform as scalars. A plane wave $$e^{imvx/\hbar}$$ will be seen as $$e^{im(v+U) x/\hbar}$$, but I would not call this a Doppler shift.
Thw weird transformation occurs because the Schroedinger wavefunction is related to the Klein-Gordon equation solution (which is a Lorentz scalar) by $$\psi_{\rm KG}(x,t) = e^{-imc^2 t} \psi_{\rm Sch}(x,t)$$ and the $$c^2$$ in the exponential cancels the factors of $$1/c^2$$ in $$t'= \gamma (t+ U x/c^2),$$ when you charge to the moving frame.