So why isn't that transformation included in it's symmetry group?
Who says it isn't? In the quoted wiki article we read:
Such a transformation is an invertible mapping of the ambient space which takes the object to itself, and which preserves all the relevant structure of the object.
Most generally the symmetry group of an object is just a set of all auto isomorphisms of that object. The word "relevant" is crucial here and in different context it means different things.
If you look at the square as a set then isomorphism = bijection and your example is valid. If you look at it as a topological space then isomorphism = homeomorphism and your example fails cause switching 2 points is not continuous. If you look at it as a subset of some vector space then isomorphism may mean linear isomorphism. If it is metric space then it may mean isometry and so on and so on...
I believe the concept you are trying to express is the one of group action: given a group $G$ and a set $X$, we say that $G$ acts on $X$ if there is a homomorphism $G\longrightarrow S(X)$, where $S(X)$ is the group of bijections $X\longrightarrow X$.
To view it better, this is the same as considering a map $\varphi:G\times X\longrightarrow X$ such that its restriction $\varphi_g:=\varphi|(\{g\}\times X)$ is a bijection for all $g\in G$, and group properties are respected: namely, $\varphi_1$ is the identity on $X$ and $\varphi :(gh,x)\longmapsto (g,(h,x))$ (or, equally, $\varphi_{gh}(x)=\varphi_g\circ\varphi_h(x)$ for all $x\in X$).
Now, the most natural action is probably the one you pointed out: we can look at $G$ as acting on itself by (left) translation, namely taking $\varphi:G\longrightarrow S(G)$ given by $g\longmapsto L_g$ for all $g\in G$, where each $L_g:G\longrightarrow G$ sends $h\in G$ to $gh$ (it is straightforward to prove this is actually an action).
The point is, and this is Cayley's theorem, this action is indeed an isomorphism $G\simeq \text{im}(\varphi)<S(G)$: in other words, any group is, in the end, a group of permutations, and any group operation is, in the end, function composition.
In this representation, to stick with your example, $\mathbb{Z}$ is the group of integer translations acting on integers (hence a subgroup of $S(\mathbb{Z})$), so that the element $n$ is actually the translation by $n$ given by $x\longmapsto n+x$ for all $x\in\mathbb{Z}$.
Best Answer
According to Schwartzman's The Words of Mathematics:
So the root of the word concerns how multiple things share the same measurement. This concept is itself symmetric under exchanging the things being measured. There is no preference for transforming $x\mapsto y$ over the reverse direction $y\mapsto x$; both should be equally possible.