Q:

Does a signal and its autocorrelation ever have the same Fourier spectra?

My understanding is that Fourier power spectra give the frequencies and corresponding intensities/amplitudes contained within a signal. In certain techniques, such as molecular dynamics, it is common to take the autocorrelation of a time-series before calculating the Fourier spectrum.
I have read that in molecular dynamics this practice can be due to the appearance of autocorrelations in quantum mechanics, and I have also read that taking the autocorrelation removes the dependence on the initial conditions.
a) Does the autocorrelation contain the same frequency components as the original signal?
b) Are there any conditions under which the Fourier power spectrum of the signal and it's autocorrelation become mathematically identical? eg. infinite signal length, perhaps.
c) What are the factors that cause these two spectra to differ?
d) Are there any other reasons to favor using the autocorrelation in place of the original signal when computing the power spectrum?
Thanks.

A:

(a). Yes, a signal and its autocorrelation have the same frequencies. If $X(f)$ is the Fourier transform of the signal and $A(f)$ the Fourier transform of its autocorrelation, then $A(f) = |X(f)|^2$. So, in general, a signal and its autocorrelation don’t have the same Fourier transform, but there is at least one exceptional case:  $x(t) = W \operatorname{sinc}(Wt)$ whose Fourier transform is $X(f) = \operatorname{rect}\left(\frac fW\right)$ and so $A(f) = |X(f)|^2 = X(f)$. Those who wish to generalize this notion to signals whose Fourier spectra are the sum of non-overlapping $\operatorname{rect}$ functions should write out the details for themselves.
All of the above is for finite-energy signals. 
If one is thinking of periodic finite-power signals with Fourier series $x(t) = \sum_{n=-\infty}^\infty c_n \exp(j2\pi nf_0t)$, then the periodic autocorrelation function has Fourier series $\sum_{n=-\infty}^\infty |c_n|^2 \exp(j2\pi nf_0t)$. Fourier series are represented by impulse trains in the frequency domain:
\begin{align}
X(f) &=  \sum_{n=-\infty}^\infty c_n \delta (f-nf_0),\\
A(f) &= \sum_{n=-\infty}^\infty |c_n|^2 \delta (f-nf_0)
\end{align}
and if we wish to continue to use the formula $A(f) = |X(f)|^2$ with the spectra of these periodic signals, then we must assume that
$$c_n \delta(f- nf_0)c_m^* \delta^*(f- mf_0) = \begin{cases}
|c_n|^2 \delta(f- nf_0), & \text{if}~m=n,\\
0, &\text{if}~m\neq n.
\end{cases}$$
Note that we are assuming that $|\delta(f-nf_0)|^2 = \delta(f-nf_0)$ and $\delta(f- nf_0) \delta^*(f- mf_0)=0$ for $m\neq n$, but if we are willing to swallow this bald and unconvincing narrative, then everything is hunky-dory.  Once again, we see that the signal and its autocorrelation have the same frequencies but that their Fourier transforms are generally not equal unless it so happens that $|c_n|^2 = c_n$ for all $n$. Note that this means that for each $n$, $c_n$ must equal $1$ or $0$. Thus, the only real-valued periodic signals whose Fourier transforms equal the Fourier transform of their autocorrelation function are of the form
$a_0 + \sum_{n=1}^\infty a_n 2\cos(2\pi f_0 t),~~ a_i \in \{0,1\}, i = 0, 1, 2, \ldots$

