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?
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 $G=S_n$.