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?


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 $Sym(G)$.
