Anisotropic scaling is a linear transformation on $\Bbb R^n$ and thus has a well defined outermorphism on $\mathcal G^n$. So if $S$ is your scaling function and $A = a_1 \wedge a_2 \wedge \cdots \wedge a_k$ is your $k$-blade, then $S(A) = S(a_1)\wedge S(a_2)\wedge \cdots \wedge S(a_k)$.