Let $k$ be an algebraically closed field of characteristic $p$, and $X$ be the plane quartic curve defined by $$x^3y+y^3z+z^3x=0,$$ which is the so-called Klein quartic. Hartshorne claims in exercise IV.5.7(b) of his book <Algebraic Geometry> that
Assume $p\ne3$, then the group $\mathrm{Aut}\, X$ is the simple group of order $168$, whose order is the maximum $84(g-1)$ allowed by (Ex. 2.5).
For the case $p\ne2,3,7$, I found a proof in e.g. Perng.pdf. A sketch: we can find $3$ special automorphisms represented by matrices $$S=\begin{pmatrix} \zeta&0&0\\ 0&\zeta^4&0\\ 0&0&\zeta^2 \end{pmatrix},\ T=\begin{pmatrix} 0&1&0\\ 0&0&1\\ 1&0&0 \end{pmatrix},\ U=\frac{1}{\sqrt{-7}}\begin{pmatrix} \zeta-\zeta^6&\zeta^2-\zeta^5&\zeta^4-\zeta^3\\ \zeta^2-\zeta^5&\zeta^4-\zeta^3&\zeta-\zeta^6\\ \zeta^4-\zeta^3&\zeta-\zeta^6&\zeta^2-\zeta^5 \end{pmatrix},$$ where $\zeta$ is a primitive $7$-th root of unity (in $k$), and the group $G$ generated by $S,T,U$ is a simple group of order $168$, achieving the Hurwitz's bound, so $G=\mathrm{Aut}\,X$.
However, this proof fails for $p=2,3,7$. For $p=2,3$, the Hurwitz's bound $84(g-1)$ can fail. For $p=7$, $X$ has a double point $[1,2,4]$ (and no other singularities), and there is no $7$-th root of unity.
I want to solve this exercise for $p=2,7$. Also, I am curious about the case $p=3$, in which the Hurwitz's bound can fail and $\mathrm{Aut}\,X$ is probably of order $>168$. Any hint or reference?
Edit: I have recently found a paper On Certain Curves of Genus Three with Many Automorphisms, which solved the case $p=2,3$. For $p=3$ the group $\mathrm{Aut}\,X$ is just $PSU(3,3^2)$, the simple group of order $6048$. However, the singular case $p=7$ is still unsolved.