7. In a three-dimensional Cartesian coordinate system, the set of points $S=\{(x, y, z) \mid x, y, z \in\{0,1\}\}$, a bijection $f: S \rightarrow S$ satisfies $|A f(A)|=1$ for any $A \in S$, then the number of such bijections $f$ is $\qquad$.

Translating the problem statement into English, the result is as follows:

7. In a three-dimensional Cartesian coordinate system, the set of points $S=\{(x, y, z) \mid x, y, z \in\{0,1\}\}$, a bijection $f: S \rightarrow S$ satisfies $|A f(A)|=1$ for any $A \in S$, then the number of such bijections $f$ is $\qquad$.
Let $S_{1}=\left\{A_{1}(0,0,0), A_{2}(0,1,1), A_{3}(1,1,0), A_{4}(1,0,1)\right\}, S_{2}=\left\{B_{1}(0,0,1), B_{2}(0,1,0), B_{3}(1,0,0), B_{4}(1,1,1)\right\}$,

then any function $f$ that meets the conditions must map all points in $S_{1}$ to $S_{2}$, and all points in $S_{2}$ to $S_{1}$.
We first find the number of mappings from $S_{1}$ to $S_{2}$ that meet the conditions. $f\left(A_{1}\right)=B_{1}, B_{2}$ or $B_{3}$, by symmetry, let's assume $f\left(A_{1}\right)=B_{1}$ first. In this case, if $f\left(A_{2}\right)=B_{2}$, then we have $\left\{\begin{array}{l}f\left(A_{3}\right)=B_{3} \\ f\left(A_{4}\right)=B_{4}\end{array}\right.$ or $\left\{\begin{array}{l}f\left(A_{3}\right)=B_{4} \\ f\left(A_{4}\right)=B_{3}\end{array}\right.$; if $f\left(A_{2}\right)=B_{4}$, then we have $\left\{\begin{array}{l}f\left(A_{3}\right)=B_{2} \\ f\left(A_{4}\right)=B_{3}\end{array}\right.$. Therefore, there are 3 mappings in this case, similarly, there are 3 mappings each when $f\left(A_{1}\right)=B_{2}$ and $f\left(A_{1}\right)=B_{3}$.

Thus, there are 9 mappings from $S_{1}$ to $S_{2}$ that meet the conditions, and similarly, there are 9 mappings from $S_{2}$ to $S_{1}$ that meet the conditions. By the multiplication principle, $f$ has a total of $9 \times 9=81$ mappings.