Solution:

The answer is $\mathbf{(E)}$. The condition that $n$ and $n+5$ have the same color is equivalent to restricting the problem to the 5 congruence classes modulo 5 (which we will keep calling $0,1,2,3,4$).

Let us now study the second condition; let us write down all the possible ways to obtain a number as the product of two others:

$0 \equiv 0 \cdot k$ with $k \in \{0,1,2,3,4\}$

$1 \equiv 1 \cdot 1 \equiv 2 \cdot 3 \equiv 4 \cdot 4$

$2 \equiv 1 \cdot 2 \equiv 3 \cdot 4$

$3 \equiv 1 \cdot 3 \equiv 2 \cdot 4$

$4 \equiv 1 \cdot 4 \equiv 2 \cdot 2 \equiv 3 \cdot 3$

Each congruence implies that if the product is white then one of the two factors is white. One can see that the color of 0 does not affect the colors of the other classes.

On the other hand, if 1 is white, then 4 is also white; if 4 is white then both 2 and 3 are white; if 2 (or 3) is white then either 3 (or 2) is white or 4 is white, but 4 white implies both 2 and 3 white.

Summing up, if 1 is white, then 2, 3 and 4 are also white. If 4 is white, but 1 is not, 2 and 3 are white. Finally if 2 is white, but neither 1 nor 4 is, then 3 is also white, and vice versa.

We therefore have four cases for the classes in $\{1,2,3,4\}$ colored white: $\{1,2,3,4\}$, $\{2,3,4\}$, $\{2,3\}$, $\emptyset$.

Recalling that 0 can be colored independently, we obtain $6 \cdot \left(1+5+5^2+5^4\right) = 3936$ different colorings.