Let $p\in\mathbb{N}$. Let $f:I\to\mathbb{R}$ differentiable in the closed interval $I$ (bounded or not), with $f(I) \subset I$, and let $g = f\circ f\circ \cdots \circ f = f^p$, where $\circ$ means composition. If $\lvert g'(x) \rvert \leq c < 1, \forall x\in I$ where $c$ is a constant, show that there is only one $a\in I$ such that $f(a) = a$.
Some thoughts
I proved it for $p=1$, i.e. $g = f$ and $\lvert g'(x) \rvert = \lvert f'(x) \rvert \leq c < 1$. Notice that $I$ can be unbounded so the existence of the fixed point is not trivial.
For the task, I use two results:
Let $f:I\to\mathbb{R}$ differentiable in the interval $I$. The function $f$ satisfies $\lvert f(x) - f(y) \rvert \leq c\lvert x - y \rvert$ for all $x,y\in I$ if and only if $\lvert f'(x) \rvert \leq c$ for all $x\in I$.
Let $0\leq c < 1$. If the sequence $(x_n)$ is such that $\lvert x_{n+2} - x_{n+1}\rvert \leq c\lvert x_{n+1} - x_n\rvert $ for all $n\in \mathbb{N}$, then the sequence $(x_n)$ converges.
Now take any point $x_0\in I$ and build the following sequence: $x_1 = g(x_0), x_2 = g(x_1), \cdots, x_n = g(x_{n-1}), \cdots$. Since $\lvert g'(x) \rvert \leq c$, from result 1 we have $\lvert g(x_{n+1}) - g(x_{n}) \rvert \leq c\lvert x_{n+1} - x_{n} \rvert$, then $\lvert x_{n+2} - x_{n+1} \rvert \leq c\lvert x_{n+1} - x_{n} \rvert$. Since $0\leq c < 1$, from result 2 we have $(x_n)\to a$. Since $I$ is closed, $a\in I$. Finally, since $x_n = g(x_{n-1})$, this means $g(a) = a$. The uniqueness comes from supposing another $b$ and applying IVT to get contradiction.
I am having trouble trying to prove for $p > 1$.
Update: I marked Matthew's suggestion because it contained the key idea for the solution. The remaining bits for the solution are in the comments to his suggestion.
Assuming your proof of the $p=1$ case is valid, you can apply it to $g=f^p$. You know $g$ has a unique fixed point $a$. You can show that $f(a)$ is a fixed point of $g$, too. But this means $a=f(a)$.