3. Determine how many functions $f:\{1,2,3,4,5,6,7\} \rightarrow\{1,2,3,4,5,6,7\}$ there are such that for all $x \in\{1,2,3,4,5,6,7\}$,

$$
f(f(x))=x \text {. }
$$
3. Let $A=\{1,2,3,4,5,6,7\}$, and let $f$ be a function that satisfies the given condition. If for some $c$ we have $f(c)=c$, then we have $f(f(c))=f(c)=c$, so the condition of the problem is satisfied for $c$. If for some $a$ and $b, a \neq b$, we have $f(a)=b$, then from the given condition we get that it must be $f(b)=f(f(a))=a$; then we also have $f(f(b))=f(a)=b$, so the condition of the problem is satisfied for $a$ and $b$. Now it is clear that the function $f$ satisfies the given conditions if and only if: for some elements $c$ of the set $A$ we have $f(c)=c$, and the remaining elements can be divided into pairs $\{a, b\}$ such that $f(a)=b$ and $f(b)=a$.

Thus, the number of elements in the set $A$ for which $f(c)=c$ is 7, 5, 3, or 1. If there are 7, then we have one function $f$; if there are 5, then the remaining pair can be chosen in $\binom{7}{2}=21$ ways; if there are 3, then the remaining pairs can be chosen in $\frac{1}{2}\binom{7}{2}\binom{5}{2}=\frac{1}{2} \cdot 21 \cdot 10=105$ ways (the first pair can be chosen in $\binom{7}{2}$ ways, the second in $\binom{5}{2}$ ways, and we divide by 2 because the order of choosing these pairs does not matter); if there is 1, then the remaining pairs can be chosen in $\frac{1}{3!}\binom{7}{2}\binom{5}{2}\binom{3}{2}=\frac{1}{6} \cdot 21 \cdot 10 \cdot 3=105$ ways (the first pair can be chosen in $\binom{7}{2}$ ways, the second in $\binom{5}{2}$ ways, the third in $\binom{3}{2}$ ways, and we divide by 3! because the order of choosing these pairs does not matter).

Therefore, the number of the desired functions is: $1+21+105+105=232$.