Let $b_i = \frac{a_{i+1}}{a_i}$ ($1 \le i \le 8$). Then for each $\{a_n\}$ satisfying the given condition, we have
$$\prod_{i=1}^{8} b_i = \prod_{i=1}^{8} \frac{a_{i+1}}{a_i} = \frac{a_9}{a_1} = 1, \text{ with } b_i \in \{2, 1, -\frac{1}{2}\} (1 \le i \le 8). \qquad \textcircled{1}$$
Conversely, a sequence of eight terms $\{b_n\}$ satisfying 1 can uniquely determine a sequence $\{a_n\}$ in the question.

In each $\{b_n\}$, there are obviously even number of $-\frac{1}{2}$ and the same number of $2$, with the remainder being $1$. Or, in other words, the numbers of $-\frac{1}{2}$ and $2$ are both $2k$, while the number of $1$ is $8-4k$. Here, it is easy to check that $k$ can only be $0, 1, 2$. Once $k$ is given, there are $C_8^{2k} C_{8-2k}^{2k}$ ways to construct $\{b_n\}$.

Therefore, the total number of $\{b_n\}$ satisfying ① is
$$N = 1 + C_8^2 C_6^2 + C_8^4 C_4^4 = 1 + 28 \times 15 + 70 \times 1 = 491.$$
The answer is 491.