Showing that $(A_t^*)$ is a martingale does not really help you in understanding the distribution of $A_T^*$. Instead, the key is your previous question.
Under $\mathbb{Q}\sim\mathbb{P}$, you have
\begin{align*}
S_t=S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right).
\end{align*}
Under $\mathbb{Q}_S\sim\mathbb{Q}$, you have
\begin{align*}
S_t^{(2)}=S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)t+\sigma W_t^{(2)}\right).
\end{align*}
From your previous question, recall that if $(W_t)$ is a standard Brownian motion under $\mathbb{Q}$, then
\begin{align*}
\hat{W}_t=W_t-\sigma t
\end{align*}
is a standard Brownian motion under the stock measure $\mathbb{Q}_S$, i.e. $\hat{W}_t\overset{d}{=}W_t^{(2)}$.
Consider now
\begin{align*}
A_T^* &= \frac{S_0}{S_T}A_T \\
&= \exp\left(-\left(r-\frac{1}{2}\sigma^2\right)T-\sigma W_T\right) \frac{1}{T}\int_0^T S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right) \mathrm{d}t \\
&= \frac{1}{T}\int_0^T S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)(t-T)-\sigma (W_T-W_t)\right) \mathrm{d}t.
\end{align*}
Following the steps of the answer to your previous question, you can manipulate the integrand to
\begin{align*}
A_T^*&= \frac{1}{T}\int_0^T S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)(T-t)+\sigma \hat{W}_{T-t}\right) \mathrm{d}t\\
&= -\frac{1}{T}\int_{T}^{0} S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)u+\sigma \hat{W}_{u}\right) \mathrm{d}u \\
&= \frac{1}{T}\int_{0}^{T} S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)u+\sigma \hat{W}_{u}\right) \mathrm{d}u.
\end{align*}
