16. Given that the function $f(x)$ is defined on $\mathbf{R}^{+}$, and satisfies the following conditions:
A. $f(x)$ is monotonically decreasing on $\mathbf{R}^{+}$, and $f(x)>\frac{1}{x^{2}}$;
B. On $\mathbf{R}^{+}$, it always holds that $f^{2}(x) f\left(f(x)-\frac{1}{x^{2}}\right)=f^{3}(1)$.
(1) Find the function value $f(1)$; (2) Provide a function $f(x)$ that satisfies the given conditions.
16. (1) When $x=1$, $\left.f^{2}(1) f(f(1)-1)\right)=f^{3}(1)$. Also, $f(1)>1$, so $f(f(1)-1)=f(1)$. Since $f(x)$ is decreasing, $f(1)-1=1$, which means $f(1)=2$;
(2) Suppose $f(x)=\frac{a}{x^{2}}$, then $\frac{a^{2}}{x^{4}} \cdot f\left(\frac{a}{x^{2}}-\frac{1}{x^{2}}\right)=2^{3}$, which simplifies to $a^{3}-8 a^{2}+16 a-8=0$, yielding $a=2$ or $a=-3 \pm \sqrt{5}$. Since $f(1)=2$, we have $a=2$, so $f(x)=\frac{2}{x^{2}}$. When $0<x_{1}<x_{2}$, $\frac{2}{x_{1}^{2}}>\frac{2}{x_{2}^{2}}$, i.e., $f\left(x_{1}\right)>f\left(x_{2}\right)$, thus $f(x)$ is monotonically decreasing. Also, $f(x)=\frac{2}{x^{2}}>\frac{1}{x^{2}}$, $f^{2}(x) f\left(f(x)-\frac{1}{x^{2}}\right)=\left(\frac{2}{x^{2}}\right)^{2} f\left(\frac{2}{x^{2}}-\frac{1}{x^{2}}\right)=\frac{4}{x^{4}} f\left(\frac{1}{x^{2}}\right)=\frac{4}{x^{4}} \cdot \frac{2}{\left(\frac{1}{x^{2}}\right)^{2}}=8=f^{3}(1)$, satisfying conditions $A$ and $B$, hence $f(x)=\frac{2}{x^{2}}$.