Auto-correlation function, an inverse problem $x[n]$ is a complex function $n=0,1,2,\cdots,L-1 $
we assume $x[n]$ is periodic in its index:  $x[n+L]=x[n]$
Its auto-correlation function $C[n]$ is uniquely defined as:
$$
C[n]=\sum_{i=0}^{L-1} x[i+n]x^*[i]
$$
$C[n]$ also has the periodic property: $$C[n+L]=C[n]\tag{1}$$
And ''conjugate-symmetry'' property: $C[-n]=C^*[n] \tag{2}$

Now my question is:
For given $C[n]$, which satisfies property (1) and (2):
Can we find the corresponding $x[n]$ ?
If yes, is it unique? 
$\qquad $ if unique, what is the method to find $x[n]$?
$\qquad $ if not unique, what is the class of those $C[n] \rightarrow \{x[n]\}$
If no, what other constraint properties should we add to $C[n]$, in order to make it yes?


No, it is not unique.
Consider the real sequences $x_0 = 0, x_1 =1$ and $y_0=1, y_1=0$ (in both cases $L=2$).
Then for $x$:
$$C_0 = x_0^2+x_1^2 =1,$$
$$C_1 = x_1x_0 + x_0x_1 = 0,$$
$$C_2 = x_0^2+x_1^2 = 1,$$
$$C_3 = x_1x_0 + x_0x_1 = 0...$$
While for $y$:
$$C_0 = y_0^2+y_1^2 =1,$$
$$C_1 = y_1y_0 + y_0y_1 = 0,$$
$$C_2 = y_0^2+y_1^2 = 1,$$
$$C_3 = y_1y_0 + y_0y_1 = 0...$$
The two sequences are shifted. In general, all shifted sequences share the same auto-correlation function. Thus, it is hard to add constraints to get a unique sequence.
