Funky function-composed within itself umpteen thousand times 
$f(x)$ is a differentiable function satisfying the following conditions:
  $$
 0 < f(x) < 1 \quad \text{for all $x$ on the interval $0 \le x \le 1$.} \\
 0 < f'(x) < 1 \quad \text{for all $x$ on the interval $0 \le x \le 1$.}
$$
  How many solutions does the equation
  $$
\underbrace{f(f(f( \ldots f}_{2016~\text{times}}(x) \ldots) =x
$$
  have on the interval $0\leq x\leq 1$?

This seems to be looking like chain rule
And since $f'$ is positive on the interval $[0,1]$ it seems to be increasing what does it do for $x>1$?
And what is the significance of 2016? 
I dont think that matters. The function is composed within itself that many times but I think maybe it doesnt matter if its 2016 or 2019 !
 A: First use induction to show that each iterate $f_n(x) = \underbrace{f(f(f( \ldots f}_{n~\text{times}}(x) \ldots)$ satisfies:
$$
\begin{align}
0 < f_n(x) < 1 \quad &\text{for all $x$ on the interval $0 \le x \le 1$,} \tag 1\\
 0 < f_n'(x) < 1 \quad &\text{for all $x$ on the interval $0 \le x \le 1$.} \tag 2
\end{align}
$$
Then use $(1)$ and the intermediate value theorem to show that $f_n(x) - x$ has at least one zero in $[0, 1]$.
Finally use $(2)$ and the mean-value theorem to show that $f_n(x) - x$ has at most one zero in $[0, 1]$.
(Remark: Instead of $0 < f'(x) < 1$ it would be sufficient to require that $|f'(x)| < 1$ on the interval.)
A: Since $f'(x)>0$ for all $x$ in the concerned interval, $f(x)$ is strictly increasing.  
So:  
$x<y \implies f(x)<f(y)$
(assuming $x,y \in [0,1]$)
Now, suppose for some $a \in [0,1],f(a) \ne a$. 
Then, either $f(a)<a$ or $f(a)>a$. 
Looking at the first case and using the fact that $f$ is strictly increasing and, crucially, that $f$ maps from $[0,1]$ back into $[0,1]$, $f(f(a))<f(a)$. Of course, by the initial assumption, $f(a)<a$, so now we get $f(f(a))<a$. We can apply this reasoning again $-f(f(f(a)))<f(a)$, but also, $f(a)<a$, so $f(f(f(a)))<a$. But if we repeat this line of reasoning an extra $2013$ times, we get that $f^{\circ 2016}(a)<a$.  
This contradicts our requirement that $f^{\circ 2016}$ fix all of $[0,1]$ so $f(a) \nless a$ for all $a \in [0,1]$.
We can reason analogously for the case where $f(a)$ is assumed to be $>a$, ruling it as an impossibility.  
So $f(x)=x$ for all $x \in [0,1]$.
But this contradicts the requirement that $f'(x)<1$ for all $0<x<1$, so no solution exists.
