II. (40 points) Given the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ satisfies $f(f(n))=2016 n-215 f(n)$.
Find $f(n)$.
Let $f^{(n)}(x)=a_{n}$. Then $a_{n}>0$, and $a_{n+2}=2016 a_{n}-215 a_{n+1}$. Hence $a_{n}-9 a_{n-1}=-224\left(a_{n-1}-9 a_{n-2}\right)$.
Let $b_{n}=a_{n}-9 a_{n-1}$. Then $b_{n}=-224 b_{n-1}=(-224)^{n-1} b_{1}$.
Let $m=-224$, we get
$a_{n}-9 a_{n-1}=m^{n-1} b_{1}$
$\Rightarrow m \cdot \frac{a_{n}}{m^{n}}-9 \cdot \frac{a_{n-1}}{m^{n-1}}=b_{1}$.
Let $c_{n}=\frac{a_{n}}{m^{n}}$, we get $m c_{n}-9 c_{n-1}=b_{1}$.
Solving for the general term, we get
$$
\begin{array}{l}
c_{n}=\left(\frac{9}{m}\right)^{n-1} c_{1}+\frac{b_{1}}{m} \cdot \frac{\left(\frac{9}{m}\right)^{n-1}-1}{\frac{9}{m}-1} \\
\Rightarrow a_{n}=\frac{9^{n}-m^{n}}{9-m} a_{1}-\frac{9^{n} m-9 m^{n}}{9-m} a_{0}>0 .
\end{array}
$$

When $n$ is even,
$$
a_{1}\frac{9^{n} m-9 m^{n}}{9^{n}-m^{n}} a_{0},
$$
as $n \rightarrow+\infty$, $a_{1} \geqslant 9 a_{0}$.
In summary, $a_{1}=9 a_{0}$, i.e., $f(n)=9 n$.