Do $|s(t)|$ and $|S(f)|$ uniquely determine $s(t)$? Consider a signal $s(t)$. My question is if you know $|s(t)|$ and $|\mathcal{FT}[s(t)](f)| = |S(f)|$ or equivalently $|s(t)|^2$ and $|S(f)|^2$ is it possible to determine $s(t)$? That is, is $s(t)$ uniquely determined by $|s(t)|$ and $|S(f)|$?

My strong feeling is that the answer is no but I can't come up with two signals $s_1(t) \neq s_2(t)$ which have the same magnitude and whose Fourier transforms have the same magnitude. Actually I can come up with trivial examples but that doesn't fully satisfy me. Let me share by work so far.

First of all, if we consider

$s_2(t) = s_1(t) e^{i\phi}$

Where $\phi$ is any real constant number then we clearly see that

\begin{align}
S_2(f) = S_1(f) e^{i\phi}
\end{align}

and

\begin{align}
|s_2(t)| &amp;= |s_1(t)|\\
|S_2(f)| &amp;= |S_1(f)|
\end{align}

So this example is a clear counterexample. However, this is just a trivial global phase that cannot be determined. That is, it is clear now that there is global phase multiplier that cannot be determined if only $|s(t)|$ and $|S(f)|$ are known. But, lets say that we either don't care about an overall phase factor or we happen to know that $s(0) = |s(0)|$ so that this global phase can be fixed.

Is it possible to come up with a non-trivial example of two functions which are a counterexample to what I am asking? I can phrase it this way.

If $|s_2(t)|= |s_1(t)|$ then

\begin{align}
s_2(t) = \Phi(t)s_1(t)
\end{align}

Where $|\Phi(t)|=1$ for all $t$. We then have that (by the convolution theorem)

\begin{align}
S_2(f) = S_1(f) \ast \Phi(f)
\end{align}

The question is is it possible for 

\begin{align}
|S_2(f)| = |S_1(f)\ast\Phi(f)| = |S_1(f)|
\end{align}

If a non-trivial $\Phi(t)$ could be demonstrated then it would be clear that $|s(t)|$ and $|S(f)|$ do not uniquely determine $s(t)$ in any sense but I can't seem to come up with anything.

If I am wrong and it turns out that $|s(t)|$ and $|S(f)|$ do in fact uniquely determine $s(t)$ then could anyone provide a method to determine $s(t)$ (up to a missing overall phase) from $|s(t)|$ and $|S(f)|$?
As pointed out in a comment by Jazzmaniac, the proof below is flawed. The mistake I made was the assumption leading to Eq. $(3)$. Of course, $|S_1(f)|=|S_2(f)|$ implies that the two spectra are related by an all-pass transformation, but the phase of that all-pass can be dependent on the input signal, and hence need not be LTI.

One example of two signals satisfying $|s_1(t)|=|s_2(t)|$ and $|S_1(f)|=|S_2(f)|$ without satisfying $(1)$ is $s_1(t)=e^{j\phi(t)}$ and $s_2(t)=e^{-j\phi(-t)}$. Clearly, $|s_1(t)|=|s_2(t)|=1$. Since $s_2(t)=s_1^*(-t)$, the corresponding spectra satisfy $S_2(f)=S_1^*(f)$, and, consequently, $|S_1(f)|=|S_2(f)|$.

The dual case would be $S_1(f)=e^{j\Phi(f)}$ and $S_2(f)=S_1^*(-f)=e^{-j\Phi(-f)}$. The signals $s_1(t)$ and $s_2(t)$ then satisfy $s_2(t)=s^*_1(t)$ and $|s_1(t)|=|s_2(t)|$.

The second example above requires the signals to be complex-valued (otherwise $s_1(t)=s_2(t)$ would hold). However, the first example also includes the real-valued case, i.e., a signal $s_1(t)$ that only attains two values: either $1$ or $-1$ (like a telegraph process). With $s_2(t)=s_1(-t)$ we guarantee $|s_1(t)|=|s_2(t)|$, and since $S_2(f)=S_1(-f)=S_1^*(f)$ we also have $|S_1(f)|=|S_2(f)|$.


OLD answer:

I think I can show that the trivial case

$$s_2(t)=e^{j\alpha}s_1(t),\qquad\alpha\in\mathbb{R}\tag{1}$$

is the only case for which $|s_1(t)|=|s_2(t)|$ and $|S_1(f)|=|S_2(f)|$ holds.

From the requirement $|s_1(t)|=|s_2(t)|$ it follows that

$$s_2(t)=e^{j\phi(t)}s_1(t)\tag{2}$$

with a real-valued function $\phi(t)$. From the requirement $|S_1(f)|=|S_2(f)|$ it follows that $s_2(t)$ can be obtained from $s_1(t)$ by all-pass filtering:

$$s_2(t)=\int_{-\infty}^{\infty}s_1(\tau)h_{AP}(t-\tau)d\tau\tag{3}$$

with $H_{AP}(f)=\mathcal{F}\{h_{AP}(t)\}$ satisfying $|H_{AP}(f)|=1$.

Since the multiplication in $(2)$ is memoryless, Equations $(2)$ and $(3)$ can only give the same result if $h_{AP}(t)=e^{j\alpha}\delta(t)$, i.e., for a trivial all-pass which just scales the input by a complex constant with magnitude equal to $1$. Consequently, the function $\phi(t)$ in $(2)$ must be constant (and equal to $\alpha$). This means that $(2)$ and $(3)$ are reduced to Eq. $(1)$.

Consequently, given $|s_1(t)|$ and $|S_1(f)|$, the signal $s_1(t)$ is determined up to a complex factor with magnitude $1$. I just do not know of any method to actually compute $s_1(t)$ from these two magnitude functions.