= Solution
Put $r=g_1=\operatorname{diag}(\alpha,\alpha^2)$ and $s=g_2$, using $\alpha^*=\alpha^{-1}=\alpha^2$. Then
$$
r^3=s^2=I_2,\qquad srs=r^{-1}.
$$
Every word can therefore be put in the form $r^k$ or $r^ks$, with $k=0,1,2$. These give six distinct <matrices>:
$$
\boxed{\begin{aligned}
G=\bigg\{&\begin{pmatrix}1&0\\0&1\end{pmatrix},
\begin{pmatrix}\alpha&0\\0&\alpha^2\end{pmatrix},
\begin{pmatrix}\alpha^2&0\\0&\alpha\end{pmatrix},\\
&\begin{pmatrix}0&1\\1&0\end{pmatrix},
\begin{pmatrix}0&\alpha\\\alpha^2&0\end{pmatrix},
\begin{pmatrix}0&\alpha^2\\\alpha&0\end{pmatrix}\bigg\}.
\end{aligned}}
$$
The <group isomorphism> $r\mapsto(123)$, $s\mapsto(12)$ identifies this <group> with the <symmetric group> $S_3$, with <permutations> composed from right to left.
Conjugation by $s$ exchanges $r,r^2$. Conjugation by $r$ cycles $s,rs,r^2s$, because $r(r^ks)r^{-1}=r^{k+2}s$. Their <conjugacy classes> are
$$
\boxed{\{I_2\},\qquad\{r,r^2\},\qquad\{s,rs,r^2s\},}
$$
with sizes $1,2,3$. There is no mixing between them: their elements have orders one, three and two, respectively.
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