(21) (12 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passes through the point $(0,1)$, and the eccentricity is $\frac{\sqrt{3}}{2}$.
(1) Find the equation of the ellipse $C$;
(2) Suppose the line $l: x=m y+1$ intersects the ellipse $C$ at points $A$ and $B$, and the point $A$ is symmetric to point $A^{\prime}$ about the $x$-axis (where $A^{\prime}$ does not coincide with $B$). Does the line $A^{\prime} B$ intersect the $x$-axis at a fixed point? If so, write down the coordinates of the fixed point and prove your conclusion; if not, explain why.
21 (1) According to the problem, we have $\left\{\begin{array}{l}b=1, \\ \frac{c}{a}=\frac{\sqrt{3}}{2}, \\ a^{2}=b^{2}+c^{2},\end{array}\right.$ Solving this, we get $a=2$, so the equation of the ellipse $C$ is $\frac{x^{2}}{4}+y^{2}=1$.
(2) From $\left\{\begin{array}{l}\frac{x^{2}}{4}+y^{2}=1, \\ x=m y+1\end{array}\right.$ eliminating $x$ yields $(m y+1)^{2}+4 y^{2}=4$, which simplifies to
$$
\left(m^{2}+4\right) y^{2}+2 m y-3=0 .
$$

Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then $A^{\prime}\left(x_{1},-y_{1}\right)$. Also, $y_{1}+y_{2}=$ $-\frac{2 m}{m^{2}+4}, y_{1} y_{2}=-\frac{3}{m^{2}+4}$.
Suppose the equation of the line passing through $A^{\prime}\left(x_{1},-y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ is
$$
y+y_{1}=\frac{y_{2}+y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right),
$$

Let $y=0$, then $x=\frac{y_{1} x_{2}+y_{2} x_{1}}{y_{2}+y_{1}}$.
Since $x_{1}=m y_{1}+1, x_{2}=m y_{2}+1$, when $y=0$,
$$
\begin{aligned}
x & =\frac{y_{1}\left(m y_{2}+1\right)+y_{2}\left(m y_{1}+1\right)}{y_{2}+y_{1}} \\
& =\frac{2 m y_{1} y_{2}+y_{1}+y_{2}}{y_{2}+y_{1}} \\
& =\frac{-\frac{6 m}{m^{2}+4}-\frac{2 m}{m^{2}+4}}{-\frac{2 m}{m^{2}+4}} \\
& =4 .
\end{aligned}
$$

Thus, the line $A^{\prime} B$ intersects the $x$-axis at the fixed point $(4,0)$.