Given,
$$x^5+10cx^3+10dx^2+5ex+f = 0$$
If there is an ordering of its roots such that,
$$\small x\_1 x\_2 + x\_2 x\_3 + x\_3 x\_4 + x\_4 x\_5 + x\_5 x\_1 - (x\_1 x\_3 + x\_3 x\_5 + x\_5 x\_2 + x\_2 x\_4 + x\_4 x\_1) = 0\tag1$$
then its coefficients are related by the quadratic in $f$,
$$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2
$$
This implies that such quintics come in pairs, having the same $c,d,e$ but differing only in $f$. An example would be the solvable *Emma Lehmer quintic*,
$$\small \color{blue}{y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y} +1=0$$
and its partner (also solvable)
$$\small \color{blue}{y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y} - (n^3 + 5n^2 + 10n + 20)(n^3 + 5n^2 + 10n + 5)/25=0$$



> 
> **Q:** Is there a subset of $n$ such that the partner also has Galois order 5? 
> 
> 
> 



On the partner of the Emma Lehmer quintic
[*Edited* to give more details]


The subset of such $n \in \bf Q$ is empty: as long as the "partner" quintic,
call it $Q$, is irreducible, its Galois group is cyclic of order $5$ over
${\bf Q}(\sqrt{5})$, but dihedral of order $10$ over $\bf Q$.


Let the roots of $Q$ be $x\_1,x\_2,x\_3,x\_4,x\_5$, ordered so that
there is a Galois automorphism $\sigma$ taking each $x\_i$ to $x\_{i+1}$
(with $x\_6 \equiv x\_1$). Then if $Q$ is irreducible over some field $K$
then its Galois group is cyclic if and only if
$$
\Delta := (x\_1 - x\_2) (x\_2 - x\_3) (x\_3 - x\_4) (x\_4 - x\_5) (x\_5 - x\_1),
$$ $$
\Delta' := (x\_1 - x\_3) (x\_3 - x\_5) (x\_5 - x\_2) (x\_2 - x\_4) (x\_4 - x\_1)
$$
are in $K$ (because $\Delta,\Delta'$ are invariant under $\sigma$, but
taken to $-\Delta,-\Delta'$ by the involution $x\_i \leftrightarrow x\_{6-i}$).
Now $\Delta \Delta' = \pm \cal D$, where
${\cal D}^2$ is the discriminant of $Q$; we compute
${\cal D} = D\_3 D'\_3 D\_4^2 / 5^3$, where
$$
D\_3 = 3 n^3 + 20 n^2 + 50 n + 50,
\quad
D'\_3 = 4 n^3 + 10 n^2 + 25 n + 25,
$$ $$
D\_4 = n^4 + 5 n^3 + 15 n^2 + 25 n + 25.
$$
Moreover, $\Delta$ and $\Delta'$ are integral over ${\bf Q}[n]$,
and even over ${\bf Z}[\frac15][n]$ with bounded denominator;
so there are not many possibilities for the factorization
${\cal D} = \Delta \Delta'$, and we can determine the correct one
by specializing $n$ and computing the $x\_i$ numerically.
We find that $\Delta$ and $\Delta'$ are
$\pm 5^{-3/2} D\_3 D\_4$ and $\pm 5^{-3/2} D\_3' D\_4$
(or vice versa if we change $x\_1,x\_2,x\_3,x\_4,x\_5$ to $x\_1,x\_3,x\_5,x\_2,x\_4$).
These are always rational over ${\bf Q}(\sqrt 5)$, but never over $\bf Q$
(for rational $n$). Therefore the Galois group is always cyclic over
${\bf Q}(\sqrt 5)$ but dihedral over $\bf Q$. $\Box$


