= Solution
Over $\mathbb F_q$, the first nonzero column of an invertible $2\times2$ <matrix> has $q^2-1$ choices and the second has $q^2-q$ choices. The surjective <determinant> homomorphism onto $\mathbb F_q^\times$ has <kernel> $\operatorname{SL}_2(q)$. Thus
$$
\boxed{|\operatorname{SL}_2(8)|=\frac{(8^2-1)(8^2-8)}{8-1}=504.}
$$
It acts on the nine one-dimensional subspaces of $\mathbb F_8^2$, namely the <projective line> $\mathbb F_8\cup\{\infty\}$, by fractional linear transformations. Any element acting trivially fixes $0,\infty,1$, which forces it to be scalar. A scalar <matrix> in $\operatorname{SL}_2(8)$ has $a^2=1$; in characteristic two this implies $a=1$. Thus the action is faithful.
The action is transitive: any nonzero vector can be completed to a <determinant>-one <basis>, so any line can be reached. The <stabilizer> of $\infty=[1:0]$ is
$$
\boxed{\left\{\begin{pmatrix}a&b\\0&a^{-1}\end{pmatrix}:
a\in\mathbb F_8^\times,\ b\in\mathbb F_8\right\}
\cong (C_2)^3\rtimes C_7.}
$$
On finite points this is $x\mapsto a^2x+ab$. Squaring permutes the seven nonzero field elements, so these are exactly the affine maps $x\mapsto ux+v$. In particular translations act transitively on all eight finite points. Thus the full action is doubly transitive. The point <stabilizer> has order $8\cdot7=56$, and all point <stabilizers> are conjugate.
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