Let's define our reversible maps as group operations applied to elements of our group for arbitrary $A \in G$ $$T_A: G \to G$$ $$T_A(x) = Ax \in G$$
The fact that maps are reversible implies that our map is bijective (as shown by proof of the Cayley's Theorem), hence a permutation.
Hence any group operation carries an element $g$ to another element of the group. Thus if we think group operations as transformations from $G$ to $G$, in the end all we got is a rearrangement of the group elements but since order doesn't matter for sets, all we got is the same set. i.e. the properties of set $G$ remains invariant under any group operation applied. And this is why group theory is indeed a study of symmetry (invariance).
As an example lets think of $(\mathbb Z, +)$. In this case we have the set $\mathbb Z = \lbrace …, -2, -1, 0, 1, 2\rbrace$. And we have infinitely many possible maps, one of them being $T_1(x) = x+1$: $-3 \to -2$, $-2 \to -1$, $-1 \to 0$, $0 \to 1$, $1 \to 2$… If we think it of as a transformation it basically provides us with just a rearrangement of the elements of $\mathbb Z$ and since order of the elements doesn't matter in a set, all we got is the same set, i.e. properties of $\mathbb Z$ are invariant (preserved) under the transformation.
Is my interpretation correct?
Best Answer
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 $x\in\mathbb{Z}$.