$S_6$ contains a subgroup $H$ that is isomorphic to $S_5$ but not conjugate to $S_5$ $$H = \langle (1 2 3 4), (3 4 5 6)\rangle $$
I'm kind of lost. 
I see that $\\S_5$ is embedded in $\\S_6$ in the form of a pentad.
Then the action of $H$ on the pentad induces a group homomorphism from $H$ into the permutations of the synthemes, which are isomorphic to $\\S_5$.
But I'm not sure how to show that the homomorphism is an isomorphism, and that is not conjugate in $\\S_5$
Let me know if I need to clarify anything.
 A: To be brief: $S_6$ has two conjugacy classes of subgroups isomorphic to $S_5$. One of them is the obvious one, consisting of stabilizers of a point. The other is the conjugacy class of your subgroup $H$. It is clear that they are not conjugate in $S_6$, because the first fixes a point, and the second one doesn't: it acts transitively. You can also think of the second one as the natural action of the group ${\rm PGL}(2,5)$.
This is connected with the fact that $S_6$ has the exceptional outer automorphism of order $2$. This automorphism interchanges the two classes of $S_5$.  
A: Clarificatory note: I had never heard the words "pentad" and "syntheme" before reading this question, and I think they might be obscure / obsolete. A Google search directed me to David Joyner's book Adventures in Group Theory, which I was lucky to have on hand.
In the language of graph theory, the pentad mentioned by the OP is a $1$-factorization of the complete graph $K_6$, i.e. a decomposition of the graph into perfect matchings; and the synthemes are the $1$-factors i.e. the perfect matchings. $S_6$ is acting on the graph via its natural action on the vertices.
Answer to question: To check that your homomorphism $H\rightarrow S_5$ is an isomorphism, you need to check 2 things: it is injective and surjective.
To check surjectivity, it is enough to know that the image of the generators of $H$ generate $S_5$. Let me label your synthemes for easy reference:
$$\begin{align*} \mathbf{1}&=\{1,2\},\{3,5\},\{4,6\}\\ \mathbf{2}&=\{1,3\},\{2,4\},\{5,6\}\\ \mathbf{3}&= \{1,4\},\{2,5\},\{3,6\}\\ \mathbf{4}&=\{1,5\},\{2,6\},\{3,4\}\\ \mathbf{5}&= \{1,6\},\{2,3\},\{4,5\}\end{align*}$$
Now $(1234)$ induces the permutation $\mathbf{1}\mapsto \mathbf{5}\mapsto\mathbf{4}\mapsto\mathbf{3}\mapsto\mathbf{1}$ of the synthemes; thus your homomorphism maps
$$(1234)\mapsto (\mathbf{1543})$$
Similarly,
$$(3456)\mapsto (\mathbf{2345})$$
Let $x=(\mathbf{1543}), y=(\mathbf{2345})$.  Playing around, you will be able to show that $x,y$ generate $S_5$. For example (I doubt this is very efficient but it seems to work), $xy=(\mathbf{152})$, and $yxy^{-1}x=(\mathbf{14325})$. (I am using the convention of seeing permutations as functions on the indices, so I am composing them right-to-left.) These two elements generate $A_5$, so the image of your homomorphism contains $A_5$. But $x,y$ themselves are odd, so it must contain all of $S_5$.
Now we have established a surjective homomorphism from $H$, a subgroup of $S_6$, to $S_5$.  Thus $H$ is a subgroup of $S_6$ of order at least $120$, and therefore index at most $6$. But we can also see that $H$ has index at least 6 from the fact that $H$ is contained in the stabilizer of the pentad you mentioned. There are 5 other pentads besides this one, and $S_6$ acts transitively on them. One can see this just by applying a few permutations to your pentad, such as $(12),(13),(14),(15),(16)$, all of which lead to different pentads. Since $S_6$ has a transitive action on at least 6 pentads, then the stabilizer of any one of them has index at least 6, and $H$, as a subgroup of one of these stabilizers, has index at least 6.
This proves $H$'s index is exactly 6, and it immediately follows that:


*

*It is an isomorphism to $S_5$;

*Actually there are exactly 6 pentads in the orbit of $S_6$'s action on the one you mentioned, and $H$ is exactly the stabilizer of this one.


This proves that $H$ is a subgroup of $S_6$ isomorphic to $S_5$.
As DonAntonio mentioned in the comments, the question of proving that $H$ is not conjugate to $S_5$ is ever so slightly ambiguous. Conjugacy is not a relation between abstract groups but a relation between subgroups of a specific group. I assume what is meant is that $H$ is not conjugate to "the usual" embedding of $S_5$ in $S_6$ as the stabilizer of one of the indices. I.e. let $A\subset S_6$ be the stabilizer of the index $6$; $A$ is naturally isomorphic to $S_5$, since it acts on the indices $1,\dots,5$. Prove that $A$ and $H$ are not conjugate in $S_6$.
As Derek Holt mentioned, this is immediate from the fact that $H$ does not stabilize any index. Since for all $a\in A$, $a(6)=6$, then given any $x\in S_6$, we have $xax^{-1}x(6)=xa(6)=x(6)$, so $xax^{-1}$ fixes $x(6)$. Thus, every element of $xAx^{-1}$ fixes $x(6)$. Since there is no index fixed by every element of $H$, $H$ cannot be of the form $xAx^{-1}$.
