Assume the quotient is $q(x)$ and the remainder is $r(x)$. Because the degree of $(x^3-x)$ is $3$, hence the degree of the remainder cannot be higher than $2$. Let $r(x)= ax^2 + bx + c$. Then $$x^{81} + x^{49} + x^{25} + x^9 + x = q(x)(x^3 - x) + r(x)$$ In order to eliminate $q(x)$, we can set the three roots of $(x^3-x)$ to the above equation, i.e. $x=-1, 0, 1$. Hence, $$ \left\{ \begin{array}{rcl} -5 &=& a - b + c\\ 0 &=&c\\ 5 &=& a+ b +c \end{array} \right. $$ Solving the above system leads to $a = c = 0, b = 5$. Therefore, hthe remainder is $r(x) = \boxed{5x}$.