= Solution
For
$$
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in GL_2(F),
$$
define its action on the <projective line> $F\cup\{\infty\}$ by the <Möbius transformation>
$$
A\cdot x=\frac{ax+b}{cx+d},
$$
with the usual conventions when the denominator vanishes or $x=\infty$. <Matrix> multiplication agrees with composition, so this is a <group action>. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero <scalar> <matrices> $Z$. The induced <group homomorphism>
$$
\phi:GL_2(F)/Z\longrightarrow S_{q+1}
$$
is therefore injective. This is the <projective general linear group action on the projective line>.
For $F=\mathbb F_4$,
$$
|GL_2(\mathbb F_4)|
=(4^2-1)(4^2-4)=15\cdot12=180.
$$
The <scalar> <subgroup> has order $|Z|=3$, so
$$
|G|=60.
$$
The four classes represented by
$$
\begin{pmatrix}1&b\\0&1\end{pmatrix},
\qquad b\in\mathbb F_4,
$$
form a <subgroup> $P$ of order four, hence a <Sylow subgroup> for the prime two. Its action is translation $x\mapsto x+b$ on $\mathbb F_4$ and fixes $\infty$. For $b\ne0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, and
$$
\boxed{\phi(P)\leq A_5}.
$$
This realizes the <Sylow 2-subgroup of PGL2 over F4>.
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