 Solution

Here the star means entrywise complex conjugation, not Hermitian conjugation. Put $s=g_3$ and $r=g_1{g}_3$. Since ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$,The generators are $g_1=rs$, $g_2=r^{2}s$ and $g_3=s$. Moving every occurrence of $s$ to the right using $sr=r^{-1}s$ reduces every word to $r^k$ or $r^{k}s$ with $k=0,1,2$. The three diagonal matrices are distinct, as are the three off-diagonal matrices, soFor an explicit group isomorphism, send $r$ to $(123)$ and $s$ to $(12)$ in the symmetric group $S_3$, composing permutations from right to left. These permutations satisfy the same relations and generate $S_3$. The normal forms above then give an isomorphism; in particular $g_1\mapsto (13)$, $g_2\mapsto (23)$ and $g_3\mapsto (12)$.

Conjugation by $s$ exchanges $r$ with $r^2$, and conjugation by $r$ cycles the three reflections. The conjugacy classes are thereforeTheir sizes are $1,2,3$, respectively. They correspond to the identity, three-cycles and transpositions in $S_3$.