Does transitive imply it's the entire symmetric group Let $G$ denote a finite group and recall that $G$ acts transitively (on itself) if and only if for all $x,y \in G$ there is a $g \in G$ such that $gx = y$.
I am wondering if transitive may imply that $G = \operatorname{Sym}{(G)}$ but then it seemed to me that the answer should be no since the action of an element $g$ is not just the transposition $x \mapsto y$. 
I then tried to think of an example that could prove to me that indeed we can have a transitive action that does not contain every permutation. I experimented with $\mathbb Z / n \mathbb Z$, for starters with $n=3$ and $n=4$. But unfortunately, it seems that if I have all transpositions $x \mapsto y$ then I can use these to build any permutation. Therefore the group action would indeed be equal to $ \operatorname{Sym}{(G)}$.

Please could someone help me resolve my confusion by explaining to me
  what my mistake in thinking is?

 A: Let $G$ be a finite group of order $n$. $G$ acts on itself by left multiplication. $g \star x=gx$. Now, this action is transitive:let $x$ and $y$ be in $G$.$y=(yx^{-1})x$. Now, if $x \in G$, $Stab(x)=\lbrace g \in G, gx=x\rbrace =\lbrace 1 \rbrace.$ This gives rise to homomorphism $ \rho:G \to Sym(G)$, and $ker \rho$ is trivial (will leave it for you to check!). So $G$ can be seen as a subgroup of $Sym(G)$, and hence, in general, if $n>2$, $n!>n$ and so $G$ acts transitively on itself, but it is not $Sym(G)$.
A: Firstly, it is relatively trivial that every group acts transitively on itself, so we don't really restrict ourselves to talking about a group acting on itself when we define transitivity. In general, we can speak of transitive actions of a group $G$ on any set $X$.
The fact that $G$ is transitive on a set $X$ does not imply $G\cong {\rm Sym}(X)$. Indeed, we can always use finite groups acting on themselves as a counterexample (assuming $|G|>2$), because the set of permutations on the group $G$ has order $|G|!$ which is much bigger than $|G|$.
It is true that transpositions generate any finite symmetric group, and transitivity implies there is always a $g$ for which $g:x\mapsto y$ for any $x,y$. But this doesn't say anything about $g$ being a transposition. Indeed, reconsider your example of $C_3$ acting on itself. Both of the nontrivial elements of $C_3$ shift elements in such a way that they have no fixed points, so none of the group elements act as a transposition. Same holds true for any group ($|G|>2$): nontrivial elements of the group act with no fixed points, hence none of them act as a transposition.
Just because you can pick an $x\in X$ and control where it is sent by picking the right $g\in G$, does not mean we can simultaneously control where other elements of $X$ are sent by the $g\in G$. If we go back to the example of a group acting on itself again, given any group element $x$, we can control where it is sent (we can send it to any $y$ by picking $g=yx^{-1}$) but the condition $x\mapsto y$ forces only one possible choice for $g$ (namely $g=yx^{-1}$ is the only group element that sends $x\mapsto y$) and so we have zero control over where any other element is sent besides $x$ itself. With a full symmetric group ${\rm Sym}(X)$ on a set $X$, we can control where each and every element of $X$ is sent off to all simultaneously and independently (insofar as we define an injective function).
Indeed, there are higher levels of transitivity. A group action on a set $X$ is called $k$-transitive if for any list $x_1,\cdots,x_k$ of $k$ distinct elements of $G$, we can control where they are all sent (i.e. for any second list $y_1,\cdots,y_k$ of $k$ distinct elements, there exists a $g\in G$ such that $g:x_1\mapsto y_1$ and so on up to $g:x_k\mapsto y_k$). Higher transitivity is fairly rare - indeed if $G\subseteq S_n$ acts $6$-transitively on the set $\{1,\cdots,n\}$ then either $G=A_n$ or $G=S_n$.
