4 added 116 characters in body

The following is wrong:

The correct answer is 240. The unnamed professor was probably thinking about the reduced automorphism group (NB. The reduced automorphism group of a hyperelliptic curve is the quotient of the automorphism group by the normal subgroup generated by the hyperelliptic involution.)

It is quite easy to describe these explicitly as well. Your curve is the unique double cover of $\mathbb P^1_{\mathbb F_5}$ which is branched over all six points which are rational over $\mathbb F_5$. The reduced automorphism group always consists of the automorphisms which preserve the unordered set of branch points, so in this case it is $PGL(2,\mathbb F_5)$. This group has order $120$.

I think I've got it now. There is a short exact sequence $1 \to \mu_2 \to \mathrm{Aut} C \to PGL(2,\mathbb F_5) \to 1$, where $\mathrm{Aut} C$ denotes the group of automorphisms defined over $\mathbb F_{25}$ or equivalently $\overline{\mathbb F}_5$.

Here is what I think is the most natural description of $\mathrm{Aut} C$ and the maps above. Let $G$ be the product $GL(2,\mathbb F_5) \times \overline{\mathbb F}_5^\ast$. Let $\Delta$ denote the normal subgroup $\{(z\cdot \mathrm{id},z^{3}) | z \in \mathbb F_5^\ast \}$ of $G$. We can let $G$ act on $x$ and $y$ via $(\gamma,\rho) \ast (x,y) \mapsto (\frac{ax+b}{cx+d},\frac{\rho y}{(cx+d)^3})$. Then $\Delta$ acts trivially, so we get an action of $G/\Delta$.

An explicit computation shows that an element $(\gamma,\rho)$ of $G$ preserves the curve $y^2 = x^5 + x$ if and only if $\det\gamma =\rho^2$. If $\Gamma$ is the subgroup of $G$ defined by this condition, then $\Delta \lhd \Gamma$ and $\mathrm{Aut} C = \Gamma/\Delta$. Now we can see in a very explicit way the short exact sequence above: the inclusion $\mu_2 \to \overline{\mathbb F}_5^\ast$ gives the first map in the short exact sequence, and the projection to the first factor gives the second map.

Moreover, it is now clear that exactly half of the automorphisms will be defined over $\mathbb F_5$, namely those for which $\det \gamma$ is a square in $\mathbb F_5$ (and this depends only on the class of $\gamma$ in $PGL(2,\mathbb F_5)$. This recovers Tim's statement that the automorphism group is an extension of $C_2$ by $A_5 \cong PSL(2,\mathbb F_5)$.

3 typo

The following is wrong:

The correct answer is 240. The unnamed professor was probably thinking about the reduced automorphism group (NB. The reduced automorphism group of a hyperelliptic curve is the quotient of the automorphism group by the normal subgroup generated by the hyperelliptic involution.)

It is quite easy to describe these explicitly as well. Your curve is the unique double cover of $\mathbb P^1_{\mathbb F_5}$ which is branched over all six points which are rational over $\mathbb F_5$. The reduced automorphism group always consists of the automorphisms which preserve the unordered set of branch points, so in this case it is $PGL(2,\mathbb F_5)$. This group has order $120$.

I think I've got it now. There is a short exact sequence $1 \to \mu_2 \to \mathrm{Aut} C \to PGL(2,\mathbb F_5) \to 1$, where $\mathrm{Aut} C$ denotes the group of automorphisms defined over $\mathbb F_{25}$ or equivalently $\overline{\mathbb F}_5$.

Here is what I think is the most natural description of $\mathrm{Aut} C$ and the maps above. Let $G$ be the product $GL(2,\mathbb F_5) \times \overline{\mathbb F}_5^\ast$. Let $\Delta$ denote the normal subgroup $\{(z\cdot \mathrm{id},z^{3}) | z \in \mathbb F_5^\ast \}$ of $G$. We can let $G$ act on $x$ and $y$ via $(\gamma,\rho) \ast (x,y) \mapsto (\frac{ax+b}{cx+d},\frac{\rho y}{(cx+d)^3})$. Then $\Delta$ acts trivially, so we get an action of $G/\Delta$.

An explicit computation shows that an element $(\gamma,\rho)$ of $G$ preserves the curve $y^2 = x^5 + x$ if and only if $\det\gamma =\rho^2$. If $\Gamma$ is the subgroup of $G$ defined by this condition, then $\Delta \lhd \Gamma$ and $\mathrm{Aut} C = \Gamma/\Delta$. Now we can see in a very explicit way the short exact sequence above: the inclusion $\mu_2 \to \overline{\mathbb F}_5^\ast$ gives the first map in the short exact sequence, and the projection to the first factor gives the second map.

Moreover, it is now clear that exactly half of the automorphisms will be defined over $\mathbb F_5$, namely those for which $\det \gamma$ is a square in $\mathbb F_5$ (and this does not depend depends only on the class of $\gamma$ in $PGL(2,\mathbb F_5)$.

The following is wrong:

The correct answer is 240. The unnamed professor was probably thinking about the reduced automorphism group (NB. The reduced automorphism group of a hyperelliptic curve is the quotient of the automorphism group by the normal subgroup generated by the hyperelliptic involution.)

I think I've got it now. There is a short exact sequence $1 \to \mu_2 \to \mathrm{Aut} C \to PGL(2,\mathbb F_5) \to 1$, where $\mathrm{Aut} C$ denotes the group of automorphisms defined over $\mathbb F_{25}$ or equivalently $\overline{\mathbb F}_5$.

Here is what I think is the most natural description of $\mathrm{Aut} C$ and the maps above. Let $G$ be the product $GL(2,\mathbb F_5) \times \overline{\mathbb F}_5^\ast$. Let $\Delta$ denote the normal subgroup $\{(z\cdot \mathrm{id},z^{3}) | z \in \mathbb F_5^\ast \}$ of $G$. We can let $G$ act on $x$ and $y$ via $(\gamma,\rho) \ast (x,y) \mapsto (\frac{ax+b}{cx+d},\frac{\rho y}{(cx+d)^3})$. Then $\Delta$ acts trivially, so we get an action of $G/\Delta$.

An explicit computation shows that an element $(\gamma,\rho)$ of $G$ preserves the curve $y^2 = x^5 + x$ if and only if $\det\gamma =\rho^2$. If $\Gamma$ is the subgroup of $G$ defined by this condition, then $\Delta \lhd \Gamma$ and $\mathrm{Aut} C = \Gamma/\Delta$. Now we can see in a very explicit way the short exact sequence above: the inclusion $\mu_2 \to \overline{\mathbb F}_5^\ast$ gives the first map in the short exact sequence, and the projection to the first factor gives the second map.

Moreover, it is now clear that exactly half of the automorphisms will be defined over $\mathbb F_5$, namely those for which $\det \gamma$ is a square in $\mathbb F_5$ (and this does not depend on the class of $\gamma$ in $PGL(2,\mathbb F_5)$.

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