15. Given that $F$ is the right focus of the ellipse $\frac{x^{2}}{1+a^{2}}+y^{2}=1(a>0)$, points $M(m, 0), N(0, n)$ are points on the $x$-axis and $y$-axis respectively, and satisfy $\overrightarrow{M N} \cdot \overrightarrow{N F}=0$. If point $P$ satisfies $\overrightarrow{O M}=2 \overrightarrow{O N}+\overrightarrow{P O}$.
(1) Find the equation of the trajectory $C$ of point $P$;
(2) Suppose a line passing through point $F$ intersects the trajectory of point $P$ at points $A, B$. The lines $O A, O B$ intersect the line $y=-a$ at points $S, T$ (where $O$ is the origin), determine whether $\overrightarrow{F S} \cdot \overrightarrow{F T}$ is a constant value. If it is, find this constant value; if not, explain the reason.
(1) $F(a, 0)$, from $\overrightarrow{M N} \cdot \overrightarrow{N F}=-m a+n(-n)=0$
$\Rightarrow m=-\frac{n^{2}}{a}$, thus $P(-m, 2 n)=\left(\frac{n^{2}}{a}, 2 n\right)$.
Therefore, the equation of the trajectory $C$ of point $P$ is $y^{2}=4 a x$.
(2) Let $A B: x=k y+a$, combining $\left\{\begin{array}{l}x=k y+a, \\ y^{2}=4 a x\end{array}\right.$
$\Rightarrow y^{2}-4 a k y-4 a^{2}=0$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
then $k_{O A}=\frac{y_{1}}{x_{1}}=\frac{4 a}{y_{1}} \Rightarrow O A: y=\frac{4 a}{y_{1}} x$,
let $x=-a \Rightarrow y_{S}=-\frac{4 a^{2}}{y_{1}}=y_{2} \Rightarrow S\left(-a, y_{2}\right)$,
similarly, we get $T\left(-a, y_{1}\right)$. Therefore, $\overrightarrow{F S} \cdot \overrightarrow{F T}=2 a \cdot 2 a+y_{1} y_{2}=0$ is a constant.