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 $336$.)

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**POSTSCRIPT**:

My thanks to Noam Elkies for the highly detailed [answer](https://mathoverflow.net/a/89931) 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.

Other than in formulas using [*Ramanujan's continued fractions*][2], I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special.  As Elkies wonderfully showed, it turns out that it is.  


  [1]: http://en.wikipedia.org/wiki/Bolza_surface
  [2]: http://sites.google.com/site/tpiezas/0015