Is it possible to construct such a function in analytical form? Suppose $f\left(f\left(x\right)\right)=\sin(x)$
Is it possible to find $f$ in closed form, or any other forms so as to visualize $f(x)$ on $x\in[-\pi,\pi]$?
Is it possible to prove the existence and uniqueness of such functions?
 A: Hint:
I guess that this is a difficult question. This functional equation is said to define the iterative square root of the RHS, of which the Babbage equation
$$f(f(x))=x$$
is the prototype.
[When a particular solution is found, such as $f(x)=-x$, more solutions are obtained as $f(x)=\phi^{-1}(-\phi(x))$, for instance $f(x)=\exp(-\ln(x))=\dfrac1x$ or $f(x)=\sqrt{-(x^2-1)+1}=\sqrt{2-x^2}$. Anyway, I don't know if a similar property holds for other RHS functions.]
You can approach this particular case by observing that $f(x)=\sin(x)$ is qualitatively close to a particular solution, as it yields a "lowered" sinusoid.

The maximum amplitude can be adjusted to $1$ by introducing a coefficient,
$$f(x)=a\sin(x)$$ so that
$$f(f(x))=a\sin(a\sin(x)).$$ With $$f\left(f\left(\frac\pi2\right)\right)=a\sin(a)=1$$we get a suitable value $$a\approx1.1141571408719$$

To get a perfect match, this function should be composed with a slightly non-linear one, on which insight can be obtained by observing the plot of $\arcsin(a\sin(a\sin(x)))$.

There is no doubt that there must exist a suitable function which is piecewise smooth. One can probably find good numerical approximations from least-squares solutions of 
$$f(f(x))=P(\sin(P(\sin(x))))=\sin(x),$$ where $P$ is a low degree polynomial of indeterminate coefficients, by identification of Taylor coefficients or by expressing equality at a number of arbitrary $x$ values.
