I stumbled upon the fact that the [Bolza surface][1] can be obtained as the locus of the equation,

$y^2 = x^5-x$

Its automorphism group has the highest order for genus $2$, namely $48$.  I recognized $x^5-x$ as a polynomial invariant of the octahedron. (In fact, the Bolza surface *is* connected to the octahedron.)  

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$y^2 = x(x^{10}+11x^5-1)$ 

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

  [1]: http://en.wikipedia.org/wiki/Bolza_surface

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**POSTSCRIPT**:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$

and,

$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$

then,

$w^5-10aw^3+45a^2w-a^2 = 0$

for arbitrary ***r***. This in fact is the *Brioschi quintic form* which the general quintic can be reduced into.  Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron.

Since the Bolza surface involved an invariant of the octahedron, I wondered if the surface using the corresponding invariant of the icosahedron would also be special.  As Elkies wonderfully showed, it turns out that it is.