2016 is absolutely awesome 18

Since $f^6(2)f^2(3)f(5)f(7) = f(20160) = 1$ we must have $f(2) = f(3) = f(5) =f(7) = 1$ .

Since $f(2) = f(2\times 1) = f(2)f(1)$ we also have $f(1) = 1$ .

If $n$ is a positive integer that is not a multiple of 2 or 5 then for some $m$ , $nm$ has last digit 3.

Therefore $f(n)f(m) = f(nm) = 1$ and hence $f(n) = 1$ .

If $n = 2^a5^bn_0$ where $n_0$ is not a multiple of 2 or 5 then

$f(n) = f^a(2)f^b(5)f(n_0) = 1$

So $f(n)$ must equal $1$ for all non-zero $n$ .

Finally note that $f(0) = f(0)f(0)$ so $f(0) = 0$ or $f(0) = 1$ .

So, there are $\boxed{2}$ functions such that all of above conditions.