1. Given $A=\{0,1,2,3,4,5,6,7\}, f: A \rightarrow$ $A$. If $i+j=7$, then $f(i) f(j)=i j$. Then, the number of mappings $f$ is $\qquad$
For $0$ and $7$ elements, since $f(i) f(j)=0$, at least one of the images of $0$ and $7$ must be $0$. There are a total of $2 \times 8-1=15$ cases.

For the images of $1$ and $6$ elements, in this case, $f(i) f(j)=6=1 \times 6=2 \times 3$, so the images of $1$ and $6$ can be: $1, 6, 6, 1, 2, 3, 3, 2$, totaling 4 cases.

Similarly, the images of $2$ and $5$ elements have 2 cases, and the images of $3$ and $4$ elements have 4 cases.
Therefore, the number of mappings $f$ is $15 \times 4 \times 2 \times 4=480$.