Can I see from the PSD if a system is a LTI system?
Suppose that the input signal to the system has power spectral density (PSD)
$S_i(f)$ while the output has PSD $S_o(f)$. Then, if the system is an LTI
system, then it must be that
$$S_o(f) = |H(f)|^2 S_i(f) \tag{1}$$
where $H(f)$ is the transfer function of the system. So, for a given pair
of PSDs, is there a $H(f)$ such that $(1)$ holds? One obvious case when
$(1)$ cannot hold for any choice of $H(f)$
is when the supports of $S_i(f)$ and $S_o(f)$ are different (in the sense
that there are frequencies $f$ for which $S_o(f) &gt; 0$ while $S_i(f) = 0$). 
So in this
case we can say unequivocally that the system is not an LTI system.

What about when the supports of $S_i(f)$ and $S_o(f)$ are identical or
the support of $S_o(f)$ is a subset of the support of $S_i(f)$ (think
bandstop or notch filter)?
Well, in this case, we do have a $|H(f)|^2$ from which we can 
deduce $|H(f)|$. Throw in a conjugate-symmetric $\angle H(f)$ and we
have constructed the transfer function $H(f) = |H(f)|e^{\angle H(f)}$
of an LTI system whose output
PSD is $S_o(f)$ when the input PSD is $S_i(f)$. Note that there can
be infinitely many LTI systems that will provide this specific
input-output relationship.



But, the above does not allow us to say that the system that actually
produced the given output PSD from the given input PSD is an LTI
system, let alone one having the specific transfer function that we have
constructed. The actual system that we are looking at might be nonlinear
or time-varying, etc. We cannot say. After all, the
$|H(f)|e^{\angle H(f)}$ that we have come up with
might be a physically unrealizable transfer
function.