Yes, this Riemann surface, call it $C: y^2 = x^{11}-11x^6-x$,
is quite special: not only does it have the maximal number of automorphisms
for a hyperelliptic surface of genus $5$, but it is a modular curve
in at least two ways, both of which exhibit its full automorphism group.

One is a classical (elliptic) modular curve of level $10$, intermediate between
$X(5)$ and $X(10)$, with $[C:X(5)] = 2$ (the hyperelliptic map) and
$[X(10):C] = 3$ (a cyclic cover); this modular curve parametrizes
elliptic curves $E$ with full level-$5$ structure and odd ${\rm Gal}(E[2])$,
or equivalently full level-$5$ structure and square $j(E)-12^3$.
Explicitly, $E$ has Weierstrass equation $Y^2 = X^3 - A(x)X/48 + B(x)/864$
where $A(x) = x^{20} + 228x^{15} + 494x^{10} - 228x^5 + 1$ and
$$
x^{30} - 522x^{25} - 10005x^{20} - 10005x^{10} + 522x^5 + 1
$$
are polynomials with roots at the $20$- and $30$-point orbits of $A_5$.
We have $A^3 - B^2 = 12^3 (x^{11}-11x^6-x)^5$,
so $j - 12^3 = B^2/(x^{11}-11x^6-x)^5$. The corresponding congruence
subgroup $\Gamma$ of ${\rm SL}_2({\bf Z})$ is the index-$2$ subgroup of
$\Gamma(5)$ consisting of matrices that reduce mod $2$ to the index-$2$
subgroup of ${\rm SL}_2({\bf Z}/2{\bf Z})$, with $[\Gamma : \Gamma(10)] = 3$.
This $\Gamma$ is normal in ${\rm SL}_2({\bf Z})$, and the quotient group
is ${\rm Aut}(C)$.

Another modular approach to $C$ is via the $(2,3,10)$ triangle group,
call it $G^*$, which appears in class VIII of the nineteen commensurability
classes tabulated in

Takeuchi, K.: Commensurability classes of arithmetic triangle groups, *J. Fac. Sci. Univ. Tokyo* **24** (1977), 201-212.

According to Takeuchi's table, $G^*$ is the normalizer of the unit-norm
group $G_1$ of a maximal order in a quaternion algebra over
${\bf Q}(\sqrt 5)$ ramified over one real place and the prime $(\sqrt 5)$.

Moreover $G_1$ is the $(3,3,5)$ triangle group, contained in $G^*$ with
index $2$. Let $G_5$ be the normal subgroup of $G_1$ consisting of
units congruent to $1 \bmod (\sqrt 5)$. Then
$G^*/G_5 \cong \lbrace \pm 1 \rbrace \times A_5$, and
the quotient of the upper half plane $\cal H$ by $G_5$ has genus $5$,
so must be our $C$. Moreover, ${\cal H} / G_5$ has no elliptic points,
so this identifies the image of the fundamental group $\pi_1(C)$ in
${\rm Aut}{\cal H} = {\rm SL}_2({\bf R})$ with an arithmetic congruence group.

P.S. Roy Smith already noted that if we allow also non-hyperelliptic
Riemann surfaces then the maximal number of automorphisms for genus $5$
is not $120$ but $192$. An explicit model for a Riemann surface $S$
with $192$ automorphisms is the intersection of three quadrics
$$
y^2 = x_0 x_1,
\phantom{and}
{y'}^2 = x_0^2 - x_1^2,
\phantom{and}
{y''}^2 = x_0^2 + x_1^2
$$
in ${\bf P}^4$. Then $(x_0:x_1:y:y':y'') \mapsto (x_0:x_1)$
gives a normal cover $S \rightarrow {\bf P}^1$ with Galois group
$N = ({\bf Z}/2{\bf Z})^3$ acting by arbitrary sign changes on
$y,y',y''$, ramified above the vertices of a regular octahedron,
with each of $x_0 x_1, x_0^2 - x_1^2, x_0^2 + x_1^2$ vanishing
on an opposite pair of vertices. I claim that there is an exact sequence
$1 \rightarrow N \rightarrow {\rm Aut}(S) \rightarrow S_4 \rightarrow 1$,
so in particular $\#({\rm Aut}(S)) = 2^3 4! = 192$. Indeed let
$G$ be the subgroup of ${\rm Aut}(S)$ that stabilizes the span of
$\lbrace x_0, x_1 \rbrace$. Then $G$ contains $N$ as the kernel of
a homomorphism $G \rightarrow {\rm Aut}({\bf P}^1)$ given by the
action on $(x_0:x_1)$. The image is contained in the
group $S_4$ of rotations of the octahedron, and indeed equals $S_4$
because any rotation permutes the three opposite pairs of vertices
and thus lifts to ${\rm Aut}(S)$. Therefore ${\rm Aut}(S)$ contains
a group $G$ of order $2^3 4! = 192$, and by the Hurwitz bound
this must be the full group of automorphisms, **QED**