Let $x=2 y$. Then, $x^{4 n}=(2 y)^{4 n}=\left\{(2 y)^{4}\right\}^{n}=\left(16 y^{4}\right)^{n}$. $y=1,2,3,4,5,6,7,8,9$ gives unit digit as 6 in $\left(16 y^{4}\right)^{n}$. But, $y=5$, gives unit digit 0 in $\left(16 y^{4}\right)^{n}$. Hence, the unit digit is 0 or 6.
Explanation as extracted from the printed page; notation may be imperfect.