jmc

We have that $$g(x^{12})=x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1.$$Note that $$(x-1)g(x)=(x-1)(x^5+x^4+x^3+x^2+x+1)=x^6-1.$$Also, $$\begin{array}{rl}g(x^{12})-6& =(x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1)-6\\ & =(x^{60}-1)+(x^{48}-1)+(x^{36}-1)+(x^{24}-1)+(x^{12}-1).\end{array}$$We can write $$(x^{60}-1)=(x^6-1)(x^{54}+x^{48}+x^{42}+\cdots +x^6+1).$$In the same way, $x^{48}-1,$ $x^{36}-1,$ $x^{24}-1,$ and $x^{12}-1$ are all multiples of $x^6-1,$ so they are multiples of $g(x).$

We have shown that $g(x^{12})-6$ is a multiple of $g(x),$ so the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$ is $\boxed{6}.$