 Solution

Over ${\mathbb{F}}_q$, the first nonzero column of an invertible $2\times 2$ 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 ${\mathrm{SL}}_2(q)$. ThusIt 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 ${\mathrm{SL}}_2(8)$ 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 $\infty =[1:0]$ isOn finite points this is $x\mapsto a^{2}x+ab$. 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 $8\cdot 7=56$, and all point stabilizers are conjugate.