What is the name of this subgroup of $S_6$? What is the name of the subgroup of $S_6$ generated by the permutations $(1\,2\,3\,4\,5\,6)$ and $(1\,4)$?
It describes the symmetries of the edges of the tetrahedron.
EDIT: It describes the symmetries of the edges of the tetrahedron up to duality, i.e. vertices are allowed to become faces and vice versa. 
I think the order of this group must be 48: Each permutation of the columns and each permutation of the entries within each column of the matrix 
$$\begin{matrix}
1 &2 &3\\
4 &5 & 6
\end{matrix}$$
describes a permutation of the group. Every permutation of the group can be arrived at in this way. For example $(1\,2\,3\,4\,5\,6)$ is arrived at like this:
$\begin{matrix}
1 &2 &3\\
4 &5 & 6
\end{matrix}\quad \overset{\text{permute columns}}{\longrightarrow} \quad \begin{matrix}
2 &3 &1\\
5 &6 &4
\end{matrix} \quad \overset{\text{permute last column}}{\longrightarrow} \quad \begin{matrix}
2 &3 &4\\
5 &6 &1
\end{matrix} $
That makes $3 ! \cdot 2! \cdot 2!\cdot 2! = 48$ permutations. 
EDIT2: As pointed out below by Derek Holt, the two permutations $(1\,2\,3\,4\,5\,6)$ and $(1\,4)$ do not actually generate the group I initially intended. 
 A: The first comment would suggest that we have the cyclic group order 6 but I don't think the symmetry group of a tetrahedron can be (isomorphic to) this. I would imagine that it would be generated by elements such as (of the same shape as) $(1,2,3)(4,5,6)$ (rotation around a vertex).
The square of $(1,2,3,4,5,6)$ is of this shape; it is $(1,3,5)(2,4,6)$ but I can't see how $(1,2,3,4,5,6)$ is in the edge symmetry group of a tetrahedron. Maybe I am missing something.
So, I think the group generated by $(1,2,3,4,5,6)$ and $(1,4),(2,5),(3,6)$ is $C_{6}$ (the latter generator being redundant) but my instinct would be that the edge symmetry group of a tetrahedron is not cyclic.
Linked to that, I can't see how $(1,2,3,4,5,6)$ is a symmetry of a tetrahedron.
One last question: Is the edge symmetry group of a tetrahedron the same as (isomorphic to) the vertex symmetry group but labelled differently? For a tetrahedron with vertices $a,b,c$ and $d$ the vertex symmetry group would be generated by elements such as $(a,b,c)$ (this is a rotation around $d$).
One last last question: Are you talking about symmetry preserving orientation? (That is what my last point assumed). If so then the symmetry group of a tetrahedron is (isomorphic to) $A_{4}$. If not then it is (isomorphic to) $S_{4}$.
By taking the edge symmetry group you pick out $S_{4}$ or $A_{4}$ (depending on if you want to preserve orientation) as a subgroup of $S_{6}$. If you label via vertices it reads straight of as $S_{4}$ or $A_{4}$.
A: As mentioned in the comment by @GerryMyerson, the $(1\,4)(2\,5)(3\,6)$ is the $3$rd power of $(1\,2\,3\,4\,5\,6)$. Hence, the subgroup generated by both is simply the subgroup generated by $(1\,2\,3\,4\,5\,6)$, an element of order $6$. Hence, the subgroup is $\mathbb{Z}_6$.
UPDATE (since OP changed the question):
The group clearly has order $12$ and is non-abelian. Hence, we have $3$ options. $A_4$,$D_6$ and $\text{Dic}_3$. Now, $A_4$ doesn't have a subgroup of order $6$. So, we have $2$ options left. Now, $(1\,4)^{-1}(1\,2\,3\,4\,5\,6)(1\,4)=(1\,5\,6\,4\,2\,3) \notin <(1\,2\,3\,4\,5\,6)>$. Hence, it has asubgroup of order $6$ that is not normal i.e. it can't be $\text{Dic}_3$.
Hence, it's the Dihedral group of order $12$ i.e. $D_6$.
EDIT: As pointed out in the comments, the group has order $24$. I'll update this answer once I can solve it. Till then, please keep in mind that the answer is wrong.
A: Let G be the group generated by $a=(1,2,3,4,5,6)$, $b=(1,4)$. 
The conjugates of $b$ by $a$ and $a^2$ give (3,6)
and (2,5) respectively. So the normal subgroup $Q$ generated by $b$ is elementary abelian of order 8 (isomorphic to $Z_2^3$) and contains $a^3=(1,4)(2,5)(3,6)$. Thus the group generated by G is a cyclic extension of $Q$ by an element of order 3, so it has order 24. 
