Find a stable transfer function $G(z)$ such that $|G(z)| = |H(z)|$
Your idea of replacing the pole $p$ outside the unit circle (i.e., $|p|&gt;1$) with a pole $1/p^*$ inside the unit circle is correct. However, as mentioned in this answer, you need to be careful with the scaling. In order to obtain the correct scaling, a factor $(z-p)$ in the denominator must be replaced by a factor $\pm |p|(z-1/p^*)$. In that way the magnitude of the transfer function will be unchanged on the unit circle, i.e., for $|z|=1$. This should also be added to the problem formulation:


  ... such that $|G(z)|=|H(z)|$ for $|z|=1$.


Clearly, the new stable transfer function $G(z)$ can be written as the original transfer function $H(z)$ cascaded with an allpass filter $A(z)$, since $|A(z)|=1$ for $|z|=1$:

$$G(z)=H(z)A(z)\quad\Longrightarrow\quad |G(z)|=|H(z)|,\quad |z|=1\tag{1}$$

For the given problem, the transfer function $A(z)$ is given by

$$A(z)=\frac{z-3}{1-3z}\tag{2}$$