We count the possible number of colorings. If four colors are used, there are two different colorings that are mirror images of each other, for a total of $2\binom{n}{4}$ colorings. If three colors are used, we choose one color to use twice (which determines the coloring), for a total of $3\binom{n}{3}$ colorings. If two colors are used, we can either choose one of those colors and color three faces with it, or we can color two faces each color, for a total of $3\binom{n}{2}$ colorings. Finally, we can also use only one color, for $\binom{n}{1}$ colorings. This gives a total of $$2\binom{n}{4}+3\binom{n}{3}+3\binom{n}{2}+\binom{n}{1}=\frac{1}{12} n^{2}\left(n^{2}+11\right)$$ colorings. Setting this equal to $n^{3}$, we get the equation $n^{2}\left(n^{2}+11\right)=12 n^{3}$, or equivalently $n^{2}(n-1)(n-11)=0$, giving the answers 1 and 11.