Amplitude Spectrum, Nyquist Frequency, mixed/min/max wavelets
Note that moving between minimum phase, maximum phase, and mixed phase involves basically two operation in the time domain:


shifting
time-reversal


If $X(e^{j\omega})$ is the Fourier transform of the sequence $x(n)$ then the transforms of the shifted and the time-reversed sequences are given by

$$x(n-k)\Longleftrightarrow e^{-jk\omega}X(e^{j\omega})\\
x(-n)\Longleftrightarrow X(e^{-j\omega})$$

Since $|e^{-jk\omega}|=1$, the magnitudes of the spectra of $x(n)$ and $x(n-k)$ are identical. Furthermore, if $x(n)$ is real-valued then $X(e^{-j\omega})=X^*(e^{j\omega})$ and consequently the spectra of $x(n)$ and $x(-n)$ also have the same magnitudes.