Let $A=\{1,2,3,5,8,9\}$. Then the number of possible functions $f:A\to A$ such that $f\left(m·n\right)=f\left(m\right)·f\left(n\right)$ for every $m,n\in A$ with $m·n\in A$ is equal to

Given,

$f\left(m·n\right)=f\left(m\right)·f\left(n\right)$

Taking $m=1$

$\Rightarrow f\left(1·n\right)=f\left(1\right)·f\left(n\right)$

$\Rightarrow f\left(1\right)=1$

Now taking $m=3 \& n=3$ we get,

$f\left(9\right)=f\left(3\right)\times f\left(3\right)$,

So, $\left(f\left(3\right), f\left(9\right)\right)$ have two possibility $\left(1,1\right) \& \left(3,9\right)$

Now taking $m=2 \& n=1$ we get,

$f\left(2·1\right)=f\left(2\right)·f\left(1\right)$

Now here $f\left(1\right)=1$, so $f\left(2\right)$ can take all $6$ numbers,

Similarly, for $f\left(5\right) \& f\left(8\right)$ there will be $6$ ways each,

So, total possible function will be, $=1\times 6\times 2\times 6\times 6\times 1=432$