8. Given the set $A=\{1,2, \cdots, 6\}$, the mapping $f: A \rightarrow A$ is a one-to-one mapping,
$$
\begin{array}{l}
f^{(2)}(x)=f(f(x)), \\
f^{(n+1)}(x)=f\left(f^{(n)}(x)\right)(x \geqslant 1) .
\end{array}
$$

Then the number of mappings $f$ that satisfy $f^{(12)}(x)=x$ for any $x \in A$ is . $\qquad$
8. 576.

Notice that there are $\mathrm{A}_{6}^{6}=$ 720 one-to-one mappings $f: A \rightarrow A$.

Consider treating $1,2, \cdots, 6$ as six points. If $f(i)=j$, then draw an arrow from $i$ to $j$.

Since $f$ is a one-to-one mapping, the resulting graph is decomposed into several cycles.

If not all $x \in A$ satisfy $f^{(12)}(x)=x$, then there is a cycle of length 5 and 1 self-loop. Thus, there are $\mathrm{C}_{6}^{1} \mathrm{~A}_{4}^{4}=144$ mappings $f$ that satisfy this condition.

Therefore, for any $x \in A$, the number of $f$ that satisfy $f^{(12)}(x)=x$ is $720-144=576$.