(Just an attempt to answer my question, I'm probably breaking all the rules, but hey I'm just a naïve mathematician at present)

Insert the Ansatz,

$$d\mu(w(S))=r^d d\mu(S)$$

Assume, that since $w_*S=\cup_i w_i(S)$ holds, where $w_*=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$, that

$$d\mu(w_*S)=r_*^dd\mu(S)=\sum_i d\mu(w_i(S))=3r^d d\mu(S)$$

$$\Rightarrow r_*^d=3r^d$$

Given that the Haussdorf measure for $r$ term shrinks the x axis, and the one for $r_*$ term stretches the y axis, which is by symmetry the same as stretching the x axis, I think its safe to assume the two constants are related by,

$$r_*=1/r$$

$$\Rightarrow 3r^d=r^{-d}$$
$$\Rightarrow 3r^{2d}=1$$
$$\Rightarrow r^{2d}=\frac{1}{3}$$
$$\Rightarrow r=3^{-\cfrac{1}{2d}}$$
$$\Rightarrow d\mu(w(S))=r^d\mu(S)=\cfrac{\sqrt{3}}{3} \cdot d\mu(S)$$
$$\Rightarrow \int_{w(S)} dH^d=\cfrac{\sqrt{3}}{3} \cdot \int_S dH^d$$
$$\Rightarrow H^d(w(S))=\cfrac{\sqrt{3}}{3}H^d(S)$$
$$\Rightarrow \cfrac{H^d(S_*)}{H^d(S)}=\cfrac{\sqrt{3}}{3}$$