So many functions

$\begin{aligned} f(x) & = ax+b \\ f(f(0)) & = 0 \\ f(a(0)+b) & = 0 \\ f(b) & = 0 \\ ab + b & = 0 \\ (a+1)b & = 0 \end{aligned}$

$\implies \begin{cases} b = 0 \\ a = -1 \end{cases}$

If $b=0$ , then $f(x) = ax$ , then we have $f(f(f(4))) = f(f(4a)) = f(4a^2) = 4a^3 = 9$ , $\implies a = \sqrt[3]{\frac 94}$ not an integer.

Therefore, the solution is $a=-1$ and $f(x) = b-x$ , and we have $f(f(f(4)) = f(f(b-4)) = f(b-b+4) = f(4) = b-4 = 9$ , $\implies b = 13$ and $f(x) = 13-x$ .

Now we have

$\begin{aligned} f(f(f(f(10)))) & = f(f(f(13-10))) \\ f(f(f(3))) & = f(f(13-3)) \\ f(f(10)) & = f(13-10) \\ f(3) & = 13-3 = \boxed{10} \end{aligned}$

if $f(f(0))=0$ , then $f(x)$ must either have reflective symmetry about the line $y=x$ , or be of the form $f(x)=kx$ for some $k$ .

Case 1:

If there is reflective symmetry about $y=x$ , then $f(f(x))=x$ . Therefore $f(f(f(f(x))))=f(f(x))=x$ , so $f(f(f(f(10))))=10$ .

Case 2:

if $y=kx$ , then $f(f(f(x)))=k^{3}x$ . The condition $f(f(f(4)))=9$ implies $4k^{3}=9$ . However, this means $k$ is not an integer so we reject this case.

Thus $f(f(f(f(10))))=\boxed{10}$