  Let $\mathbb{K}$ be the algebraic closure of a finite field $\mathbb{F}_q$ of
odd characteristic $p$. For a positive integer $m$ prime to $p$, let
$F=\mathbb{K}(x,y)$ be the transcendency degree $1$ function field defined by
$y^q+y=x^m+x^{-m}$. Let $t=x^{m(q-1)}$ and $H=\mathbb{K}(t)$. The extension
$F|H$ is a non-Galois extension. Let $K$ be the Galois closure of $F$ with
respect to $H$. By a result of Stichtenoth, $K$ has genus $g(K)=(qm-1)(q-1)$,
$p$-rank (Hasse-Witt invariant) $\gamma(K)=(q-1)^2$ and a
$\mathbb{K}$-automorphism group of order at least $2q^2m(q-1)$. In this paper
we prove that this subgroup is the full $\mathbb{K}$-automorphism group of $K$;
more precisely $Aut_{\mathbb {K}}(K)=Q\rtimes D$ where $Q$ is an elementary
abelian $p$-group of order $q^2$ and $D$ has a index $2$ cyclic subgroup of
order $m(q-1)$. In particular, $\sqrt{m}|Aut_{\mathbb{K}}(K)|> g(K)^{3/2}$, and
if $K$ is ordinary (i.e. $g(K)=\gamma(K)$) then
$|Aut_{\mathbb{K}}(K)|>g^{3/2}$. On the other hand, if $G$ is a solvable
subgroup of the $\mathbb{K}$-automorphism group of an ordinary, transcendency
degree $1$ function field $L$ of genus $g(L)\geq 2$ defined over $\mathbb{K}$,
then by a result due to Korchm\'aros and Montanucci, $|Aut_{\mathbb{K}}(K)|\le
34 (g(L)+1)^{3/2}<68\sqrt{2}g(L)^{3/2}$. This shows that $K$ hits this bound up
to the constant $68\sqrt{2}$.
  Since $Aut_{\mathbb{K}}(K)$ has several subgroups, the fixed subfield $F^N$
of such a subgroup $N$ may happen to have many automorphisms provided that the
normalizer of $N$ in $Aut_{\mathbb{K}}(K)$ is large enough. This possibility is
worked out for subgroups of $Q$.
