21. (12 points) Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$ pass through the point $(0,1)$, and have an eccentricity of $\frac{\sqrt{3}}{2}$.
(1) Find the equation of the ellipse $C$.
(2) Let the line $l: x=m y+1$ intersect the ellipse $C$ at points $A$ and $B$, and let the point $A$ be symmetric to point $A^{\prime}$ with respect to the $x$-axis (point $A^{\prime}$ does not coincide with point $B$). Is the line $A^{\prime} B$ intersecting the $x$-axis at a fixed point? If so, write down the coordinates of the fixed point and prove your conclusion; if not, explain your reasoning.
21. (1) According to the problem, we have
$$
\left\{\begin{array}{l}
b=1, \\
\frac{c}{a}=\frac{\sqrt{3}}{2}, \\
a^{2}=b^{2}+c^{2}
\end{array} \Rightarrow a=2 .\right.
$$

Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+y^{2}=1$.
(2) From $\left\{\begin{array}{l}\frac{x^{2}}{4}+y^{2}=1, \\ x=m y+1\end{array}\right.$
$$
\Rightarrow\left(m^{2}+4\right) y^{2}+2 m y-3=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $A^{\prime}\left(x_{1},-y_{1}\right)$, and
$$
y_{1}+y_{2}=-\frac{2 m}{m^{2}+4}, y_{1} y_{2}=-\frac{3}{m^{2}+4} \text {. }
$$

Let the equation of the line passing through points $A^{\prime}\left(x_{1},-y_{1}\right), B\left(x_{2}, y_{2}\right)$ be
$$
\begin{array}{l}
y+y_{1}=\frac{y_{2}+y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right) . \\
\text { Let } y=0 . \text { Then } \\
x=\frac{y_{1} x_{2}+y_{2} x_{1}}{y_{2}+y_{1}} . \\
=\frac{y_{1}\left(m y_{2}+1\right)+y_{2}\left(m y_{1}+1\right)}{y_{2}+y_{1}} \\
=\frac{2 m y_{1} y_{2}+y_{1}+y_{2}}{y_{2}+y_{1}}=4,
\end{array}
$$

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