On the partner of the Emma Lehmer quintic 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? 

 A: [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 $\bf Q$.  $\Box$
