5.2.

From the problem, we know that $a * b$ represents the remainder of $a b$ modulo 5.
Let $a=0, b=2$, we get $f(0)=0$.
Since $f$ is a bijection, we have
$$
\begin{array}{l}
f(1 * 2)=f(2)=f(1) * f(2), \\
f(1 * 3)=f(3)=f(1) * f(3) .
\end{array}
$$

Thus, $f(1) f(2) \equiv f(2) \pmod{5}$.
Since $f(2) \neq 0$, we have $f(1)=1$.
Similarly, $f(4)=4$.
Upon verification, $f(2)=3, f(3)=2$ and $f(2)=2, f(3)=3$ both satisfy the conditions.
Therefore, there are only 2 functions $f$ that meet the requirements.
5. Define the operation $a * b=a b-5\left[\frac{a b}{5}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$, and the set $A=$ $\{0,1,2,3,4\}$. A bijection $f: A \rightarrow A$ satisfies $f(a * b)=$ $f(a) * f(b)$. Then the number of such $f$ is .