I couldn't find such a system but I have also not been able to prove otherwise.
Firstly, I don't know exactly how to take the inverse Z-Transform of $z^*$.
Secondly, I don't know the ROC associated with it so there has to be more than one discrete-time function that has the Z-Transform, $z^*$.
You are correct, there is no such stable LTI system which has an impulse response with $\mathrm{Z}$-Transform = $z^*$.
Proof by contradiction:
Assume that there is an LTI system which is BIBO stable and has the transfer function: $H(z) = z^*$.
Since $H(z)$ is stable, $|z|=1$ is inside its ROC. Therefore taking $z=e^{j\omega}$,
\begin{align*} H(e^{j\omega}) & = (e^{j\omega})^* = e^{-j\omega}\\ H(e^{j\omega}) & = \text{DTFT} \{h[n]\} \end{align*}
The $\text{IDTFT}\{e^{-j\omega}\} = \delta[n-1]$.
Then $H(z) = \sum_{n=-\infty}^{\infty} h[n]z^{-n} = \delta[n-1]z^{-n} = z^{-1}$
But $z^{-1} \neq z^{*}$. This is a contradiction!
Therefore, there is no stable LTI system whose transfer function is $z^{*}$.
Note that the transfer function of a system with impulse response $h[n]$ is defined by
$$H(z)=\sum_{n=-\infty}^{\infty}h[n]z^{-n}\tag{1}$$
Eq. $(1)$ is also called the $\mathcal{Z}$-transform of $h[n]$.
If you look at $(1)$, it should be clear that there's no way you could ever create a transfer function $H(z)=z^*$. Stability is totally unrelated. There exists neither a stable nor an unstable system with transfer function $H(z)=z^*$.
In more mathematical terms, $H(z)=z^*$ is not an analytic function, and hence it can't be the $\mathcal{Z}$-transform of any sequence. Similarly, the functions $H(z)=|z|$ or $H(z)=\textrm{Re}\{z\}$ are no valid transfer functions because they are not analytic.
