9. (16 points) Given

the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$
with the left and right endpoints of the major axis being $A$ and $B$, and the left and right foci being $F_{1}$ and $F_{2}$. If there exists a point $P$ on the ellipse $\Gamma$ such that $P F_{1}$ and $P F_{2}$ trisect $\angle A P B$, among all ellipses $\Gamma$ that satisfy the above conditions, find the number of different eccentricities $e$.
9. From $S_{\triangle P A F_{1}}=S_{\triangle P B F_{2}}$, we get $P A \cdot P F_{1}=P B \cdot P F_{2}$. 
Therefore, $P$ must be the endpoint of the minor axis, let's assume point $P(0, b)$. The slopes of lines $P A$ and $P F_{1}$ are $\frac{b}{a}$ and $\frac{b}{c}$, respectively, so $\tan \angle A P F_{1}=\frac{\frac{b}{c}-\frac{b}{a}}{1+\frac{b}{c} \cdot \frac{b}{a}}=\frac{b(a-c)}{a c+b^{2}}$. Let $O$ be the origin.

From $\tan \angle O P F_{1}=\frac{c}{b}$
$\Rightarrow \tan \angle F_{1} P F_{2}=\tan 2 \angle O P F_{1}$
$$
=\frac{2 \frac{c}{b}}{1-\left(\frac{c}{b}\right)^{2}}=\frac{2 b c}{b^{2}-c^{2}} \text {. }
$$

It is easy to see that $\angle A P F_{1}=\angle F_{1} P F_{2}$, so $\frac{b(a-c)}{a c+b^{2}}=\frac{2 b c}{b^{2}-c^{2}}$.
Substituting $b^{2}=a^{2}-c^{2}$ into the above equation gives $(a-c)\left(a^{2}-2 c^{2}\right)=2 c\left(a c+a^{2}-c^{2}\right)$.
Given $e=\frac{c}{a} \in(0,1)$
$$
\Rightarrow 4 e^{3}-4 e^{2}-3 e+1=0 \text {. }
$$

Let $f(t)=4 t^{3}-4 t^{2}-3 t+1(t \in \mathbf{R})$. Then $f^{\prime}(t)=12 t^{2}-8 t-3$.
Setting $f^{\prime}(t)=0$, we get two real roots
$$
\begin{array}{l}
t_{1}=\frac{2-\sqrt{13}}{6} \in(-1,0), \\
t_{2}=\frac{2+\sqrt{13}}{6} \in(0,1) .
\end{array}
$$

Since $f\left(t_{1}\right)>0, f\left(t_{2}\right)0, f(1)<0$, the equation (1) has exactly one real root in the interval $(0,1)$.

Therefore, the number of different eccentricities $e$ that satisfy the condition is 1.