18. (14 points) Given the parabola $C_{1}$:
$$
y=a x^{2}-2 a m x+a m^{2}+2 m+1(a>0, m>0)
$$

with vertex $A$, and the vertex $B$ of parabola $C_{2}$ is on the $y$-axis, and parabolas $C_{1}$ and $C_{2}$ are centrally symmetric about $P(1,3)$.
(1) When $a=1$, find the equation of parabola $C_{2}$ and the value of $m$;
(2) Let the intersection point of parabola $C_{2}$ with the positive $x$-axis be $C$. When $\triangle A B C$ is an isosceles triangle, find the value of $a$.


18. (1) When $a=1$, we have
$$
\begin{array}{l}
y=a x^{2}-2 a m x+a m^{2}+2 m+1 \\
=(x-m)^{2}+2 m+1,
\end{array}
$$

Therefore, the vertex $A(m, 2 m+1)$.
Also, $P(1,3)$, and let the equation of line $A B$ be $y=k x+b$.
Substituting the coordinates of points $A$ and $P$ gives
$$
\left\{\begin{array}{l}
2 m+1=k m+b, \\
3=k+b .
\end{array}\right.
$$
(1) - (2) yields $2 m-2=(m-1) k$.

Since $m \neq 1$ (if $m=1$, then points $A$, $B$, and $P$ would coincide, which is not the case). Therefore, $k=2, b=1$.

Thus, the equation of line $A B$ is $y=2 x+1$, and the vertex $B$ of $C_{2}$ is $(0,1)$.

Since $C_{2}$ and $C_{1}$ are centrally symmetric about point $P$, the parabolas have the same opening size but opposite directions, so the equation of $C_{2}$ is
$$
y=-x^{2}+1.
$$

As shown in Figure 16, since points $A$ and $B$ are centrally symmetric about point $P(1,3)$, draw $P E \perp y$-axis at point $E$, and draw $A F \perp y$-axis at point $F$. It follows that $\triangle B P E \sim \triangle B A F$. Therefore, $A F=2 P E$, which means $m=2$.
(2) In the right triangle $\triangle A B F$, since $A B=\sqrt{2^{2}+4^{2}}=2 \sqrt{5}<5$, when $\triangle A B C$ is an isosceles triangle, there are only the following two cases:
(i) As shown in Figure 17 (a), if $B C=A B=2 \sqrt{5}$, then $O C=$ $\sqrt{B C^{2}-O B^{2}}=\sqrt{19}$, so $C(\sqrt{19}, 0)$.
Since $C(\sqrt{19}, 0)$ lies on $y=-a x^{2}+1$, we have $a=\frac{1}{19}$.
(ii) As shown in Figure 17 (b), if $A C=B C$, let $C(x, 0)$, and draw $A D \perp x$-axis at point $D$.
In the right triangle $\triangle O B C$, $B C^{2}=x^{2}+1$;
In the right triangle $\triangle A D C$, $A C^{2}=(x-2)^{2}+25$.
By $x^{2}+1=(x-2)^{2}+25$, we solve for $x=7$.
Since $C(7,0)$ lies on $y=-a x^{2}+1$, we have $a=\frac{1}{49}$.
In summary, the values of $a$ that make $\triangle A B C$ an isosceles triangle are two: $a_{1}=\frac{1}{19}, a_{2}=\frac{1}{49}$.
(Wang Shengyu provided)