Equation with Galois group twisted $S_{3}$ I note that a Galois group is not just a Galois group. Let $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$,  denote the roots of a quartic equation. Then $x^4-5x^2+6$ has Galois group $Z_{2}^2$, where the permutations of the roots are given by ($r$'s and fixed roots omitted) $()$, $(12)$, $(34)$, and $(12)(34)$, while $x^4+1$ also has Galois group $Z_{2}^2$, but its permutations are $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$. 
These situations are different. The first polynomial can be factored, but the second one is irreducible over the rationals. So actually it might be better to refer to a Galois group as a group acting on a set or a permutation group instead of just a group. 
In that case, I wonder if there is a quintic equation whose Galois group/set is $()$, $(123)$, $(132)$, $(12)(45)$, $(23)(45)$, $(13)(45)$. This group is $S_{3}$, but its $2$-cycles are paired up with $(45)$; I have seen this called "twisted $S_{3}$" on some sites. If so, it can't be like the usual $S_{3}$ equation such as $x^3-2$. Since this group permutes $5$ elements, it must be the group of a quintic, not a cubic, but it looks like it has cubic roots. Further, the usual cubic equation such as $x^3-2$ has a non-square discriminant (in this case, $108$), but twisted $S_{3}$ is a subgroup of $A_{5}$, which means that the quintic has to have a square discriminant. So can there be such an equation?
 A: An example
$(x^3 - 2)(x^2+3)$.
Edit: With respect to the claim that a Galois group is not just a Galois group: let $P$ be a polynomial with rational coefficients and let $r_1,\dots, r_n$ the roots of $P$. For simplicity assume that $P$ has simple roots. The field $Q^P:=\mathbb Q(r_1,\dots,r_n)$ of descomposition of $P$
is a Galois extension of $\mathbb Q$.
To the polynomial $P$ one usually attaches the group $G(P):=Gal(\mathbb Q^P/ \mathbb Q)$ and a natural representation
as a group of permutation, that is an embedding
$$
G(P) \hookrightarrow S_n
$$
into the symmetric group of $n$ elements; just by identifying   $\{r_1,\dots r_n\}$ with $\{1,\dots n\}$. 
This embedding depends strongly on $P$ but $G(P)$ weakly on $P$. 
For example let $P=x^4 + 1$, and let $r$ be a root of $P$. Then $\mathbb Q^P= Q(r)$. As the OP mentioned the group attached to $P$ is represented as a transitive permutation group by () , (12)(34), (13)(24), and (14)(23).
Since $r$ is a $8$th roof of unity it is direct by Euler's identity that $\mathbb Q^P= \mathbb Q(\sqrt 2, i)=\mathbb Q^{(x^2 -2)(x^2+1)}$. 
The permutation group attached to $Q=(x^2 -2)(x^2+1)$ is $(), (1,2),(3,4),(1,2)(3,4)$, a non-transitive permutation group. As abstract groups they are isomorphic.
I would say that a Galois group is a group of field automorphisms but a particular realization of it as a permutation group is not automatically inherent of $G$ but a consequence of the procedure used to obtain $K/\mathbb Q$. 
Number theorists have been interested in more exotic representations. For example the theory of cyclotomic fields represents 
$Gal(\mathbb Q^Q/\mathbb Q)$ canonically as the group of units $(\mathbb Z/ 8\mathbb Z)^\times$ which is more than a group of permutations. 
