I. Comparison

It doesn't seem to be well-known that the generic cubic for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed it had this property after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p=6m+1$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p = 10m+1$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Questions:

1. Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?
2. Given the known property of the Lehmer quintic that, $$x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0$$ does this require the same ordering of roots $(x_1, x_2, x_3, x_4, x_5) = (a,b,c,d,e)?$

III. Example

Let $n=-1$ and we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. We can then verify it obeys the two relations in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?