10. Solution: (1) $\because P$ is any point on the ellipse, $\therefore\left|P F_{1}\right|+\left|P F_{2}\right|=2 a$ and $a-c \leqslant\left|P F_{1}\right| \leqslant a+c$, $y=\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=\left|\overrightarrow{P F_{1}}\right|\left|\overrightarrow{P F_{2}}\right| \cos \angle F_{1} P F_{2}=\frac{1}{2}\left[\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-4 c^{2}\right]$ $=\frac{1}{2}\left[\left|P F_{1}\right|^{2}+\left(|2 a|-\left|P F_{1}\right|\right)^{2}-4 c^{2}\right]=\left(\left|P F_{1}\right|-a\right)^{2}+a^{2}-2 c^{2}$.
When $\left|P F_{1}\right|=a$, $y$ has the minimum value $a^{2}-2 c^{2}$; when $\left|P F_{1}\right|=a-c$ or $a+c$, $y$ has the maximum value $a^{2}-c^{2}$. $\therefore\left\{\begin{array}{l}a^{2}-c^{2}=3 \\ a^{2}-2 c^{2}=2\end{array},\left\{\begin{array}{l}a^{2}=4 \\ c^{2}=1\end{array}, b^{2}=a^{2}-c^{2}=3 . \therefore\right.\right.$ The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, substituting $y=k x+m$ into the ellipse equation gives $\left(4 k^{2}+3\right) x^{2}+8 k m x+4 m^{2}-12=0$.
$\therefore x_{1}+x_{2}=\frac{-8 k m}{4 k^{2}+3}, x_{1} x_{2}=\frac{4 m^{2}-12}{4 k^{2}+3}$.
$\because y_{1}=k x_{1}+m, y_{2}=k x_{2}+m, y_{1} y_{2}=k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2}$,
$\because$ the circle with diameter $M N$ passes through point $A, \therefore \overrightarrow{A M} \cdot \overrightarrow{A N}=0, \therefore 7 m^{2}+16 k m+4 k^{2}=0$,
$\therefore m=-\frac{2}{7} k$ or $m=-2 k$ both satisfy $\Delta>0$,
If $m=-2 k$, the line $l$ always passes through the fixed point $(2,0)$, which is not consistent with the problem, so it is discarded,
If $m=-\frac{2}{7} k$, the line $l: y=k\left(x-\frac{2}{7}\right)$ always passes through the fixed point $\left(\frac{2}{7}, 0\right)$.