= Solution
The identity column gives the <group order> as the sum of squared <irreducible character degrees>:
$$
\boxed{|G|=1+4\cdot7^2+8^2+3\cdot9^2=504.}
$$
By <column orthogonality>, the squared norm of a column is the corresponding <centralizer> order. In source order these orders and the resulting <conjugacy class> sizes are
$$
\boxed{|C_G(g)|=(504,8,9,9,9,9,7,7,7),}
$$
$$
|g^G|=(1,63,56,56,56,56,72,72,72).
$$
For example the fourth column has squared norm $1+1+(|\alpha|^2+|\beta|^2+|\gamma|^2)+1=9$, while the seventh has squared norm $1+1+(|\xi|^2+|\eta|^2+|\phi|^2)=7$.
An element's order divides its <centralizer> order. <Cauchy's theorem for finite groups> provides an <involution>, and only $C_2$ has an even <centralizer> order. Hence that entire class consists of <involutions>. The last three classes consist of elements of order seven. In $C_3,C_4,C_5,C_6$ the possibilities are three and nine. The three real numbers $\alpha,\beta,\gamma$ are roots of $t^3-3t+1$, which has no rational root and is irreducible over $\mathbb Q$. A real <character> value at an element of order three belongs to $\mathbb Q(e^{2\pi i/3})\cap\mathbb R=\mathbb Q$. Therefore the irrational real entries in $C_4,C_5,C_6$ rule out order three there. Cubing any of these order-nine elements produces an element of order three, which must belong to the sole remaining class $C_3$. Thus
$$
\boxed{\operatorname{ord}(C_1),\ldots,\operatorname{ord}(C_9)
=(1,2,3,9,9,9,7,7,7).}
$$
The three deductions below give nine <Sylow subgroups> for $2$ with pairwise trivial intersections. Their conjugation action is transitive. Fix one, $S$. If $T\ne S$, put $H=S\cap N_G(T)$. The product $HT$ is a $2$-<subgroup>, so maximality of $T$ gives $H\leq T$. Trivial intersection then gives $H=1$. Thus the <orbit> of $T$ under $S$ has size eight, exhausting the other eight points. The point <stabilizer> $N_G(S)$ is therefore transitive on the remaining points: \b[the action is doubly transitive].
For a doubly transitive action, its <permutation character> has the form $1+\theta$ with $\theta$ irreducible. Here $\theta(1)=8$, and the unique degree-eight row is $\chi_6$. Therefore the numbers of fixed points are the entries of
$$
\boxed{\pi=\chi_1+\chi_6=(9,1,0,0,0,0,2,2,2).}
$$
Every nonidentity element consequently fixes at most two points. This also shows that the action is faithful.
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