Your curve is hyperelliptic.
If $X_g$ is a hyperelliptic curve of genus $g$, then $\textrm{Aut}(X_g)$ is a central extension of degree $2$ of one of the groups $$\mathbb{Z}_n, D_n, A_4, S_4, A_5,$$
see Shaska - Determining the automorphism group of a hyperelliptic curve.
In the case of Bolza curve the polynomial $x^5-x$ is invariant by the automorphism group of the octahedron, which is $S_4$. In fact, the automorphism group of the Bolza curve is a central extension of $S_4$ by the group of order $2$ generated by the hyperelliptic involution, hence it has order $2 \cdot |S_4|=48$.
Regarding your curve, the polynomial at the right hand side is invariant by the automorphism group of the icosahedron, which is $A_5$. Then the automorphism group is a central extension of $A_5$ by the hyperelliptic involution, hence it has order $2 \cdot |A_5|= 120$.
