The modular curve $Y_0(5)$ parametrizes elliptic curves $E$ with an isogeny of degree five. So an element of $Y_0(5)$ can be interpreted as
$E \xrightarrow{\phi} E'$.
Suppose we are working over a finite field. An automorphism of these curves is an automorphism of the curve $E$, that sends the kernel $G$ of the map $\phi$ to itself, over some algebraic closure. The elements of $Y_0(5)$ do admit automorphisms, because any curve has the hyperelliptic involution.
Because $G$ is of order five, it can have at most four automorphisms. If the curve $E$ has six automorphisms, it is clear, by comparing the sizes of the groups, that only the identity and the hyperelliptic involution respect the subgroup $G$. If $E$ has four automorphisms, this kind of reasoning doesn't work. Is there a similar result if $E$ has four automorphisms? To me it is not clear if there are always only two automorphisms that map $G$ to itself, or that there could be four. Many thanks in advance!
Suppose that $E$ has CM by $\mathbb Z[i]$. The prime $5$ splits in $\mathbb Z[i],$ as $5 = \pi\overline\pi,$ where $\pi = 2 + i$.
Let $G$ be the subgroup of $\pi$-torsion in $E$, or the subgroup of $\overline{\pi}$-torsion. These are each of order $5$, and are preserved under mult. by all of $\mathbb Z[i]$, and in particular by the automorphism group $\{\pm 1,\pm i\}$. Conversely, if $G$ is preserved under $\{\pm 1,\pm i\}$, then in fact is preserved under all of $\mathbb Z[i]$, and so is easily checked to be either the $\pi$ or $\overline\pi$ torsion subgroup.
So of the $6$ subgroups of order $5$, two of them are preserved the full automorphism group, and the other four aren't.