Four. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$, $F_{1}$ and $F_{2}$ are its left and right foci, $P$ is any point on the ellipse $C$, the centroid of $\triangle F_{1} P F_{2}$ is $G$, the incenter is $I$, and $I G=\lambda F_{1} \boldsymbol{F}_{2}$.
(1) Find the eccentricity $e$ of the ellipse $C$;
(2) A line $l$ passing through the focus $F_{2}$ intersects the ellipse $C$ at points $M$ and $N$. If the maximum area of $\triangle F_{1} M N$ is 3, find the equation of the ellipse $C$.
(1) Let $P(x, y)(y \neq 0)$.
Since $F_{1}(-c, 0), F_{2}(c, 0)$, then $G\left(\frac{x}{3}, \frac{y}{3}\right)$.
Since $IG=\lambda F_{1} F_{2}$, then $IG \parallel F_{1} F_{2}$. Therefore, the y-coordinate of point $I$ is the same as the y-coordinate of point $G$.
Thus, the inradius of $\triangle F_{1} P F_{2}$ is $r=\frac{1}{3}|y|$.
Hence, $S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right| \cdot|y|$
$$
=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+\left|F_{1} F_{2}\right|\right) \cdot r,
$$

which means $\frac{1}{2} \cdot 2 c \cdot|y|=\frac{1}{2}(2 a+2 c) \cdot \frac{1}{3}|y|$.
Thus, $a=2 c, e=\frac{c}{a}=\frac{1}{2}$.
(2) Let the equation of $l$ be $x=m y+c$, and combine it with the equation of the ellipse $C$, eliminating $x$ and simplifying, we get
$$
\left(a^{2}+b^{2} m^{2}\right) y^{2}+2 b^{2} c m y-b^{4}=0 .
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.
Then $y_{1}+y_{2}=-\frac{2 b^{2} c m}{a^{2}+b^{2} m^{2}}, y_{1} y_{2}=\frac{-b^{4}}{a^{2}+b^{2} m^{2}}$.
Thus, $S_{\triangle F_{1} M N}=\frac{1}{2}\left|F_{1} F_{2}\right| \cdot\left|y_{1}-y_{2}\right|$
$$
=c \sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}=\frac{2 a b^{2} c \sqrt{1+m^{2}}}{c^{2}+b^{2}\left(1+m^{2}\right)} \text {. }
$$

Since $e=\frac{c}{a}=\frac{1}{2}$, we have
$$
a=2 c, b=\sqrt{a^{2}-c^{2}}=\sqrt{3} c \text {. }
$$

Thus, $S_{\triangle F_{1} M N}=\frac{12 c^{2} \sqrt{1+m^{2}}}{1+3\left(1+m^{2}\right)}$
$$
=4 c^{2} \cdot \frac{1}{\sqrt{1+m^{2}}+\frac{\frac{1}{3}}{\sqrt{1+m^{2}}}} \text {. }
$$

Let $f(t)=t+\frac{\frac{1}{3}}{t}$, where $t=\sqrt{m^{2}+1} \geqslant 1$.
It is easy to prove that $f(t)$ is an increasing function on $[1,+\infty)$, so, $f(t)_{\min }=f(1)=\frac{4}{3}$.
Thus, $S_{\triangle F_{1} M N} \leqslant 4 c^{2} \cdot \frac{1}{\frac{4}{3}}=3 c^{2}$, which means the maximum value of $S_{\triangle F_{1} M N}$ is $3 c^{2}$.
Therefore, $3 c^{2}=3$, which implies $c=1$. Thus, $a=2, b=\sqrt{3}$.
Hence, the equation of the required ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.