The single object in the groupoid corresponding to a group $G$ does not really correspond to anything in the group - but you can think of it as a thing which has as its group of symmetries the group $G$.
As an example, the category whose only object is the set $[n]=\{1,\dotsc,n\}$, and morphisms are bijections from this object onto itself. Then the symmetries of $[n]$ are given by the group $S_n$.
For a more subtle example, consider a topological space $X$, and fix a point $x\in X$. Consider the category with one object, namely the point $x$. A morphism from $x\to x$ is a homotopy-class of paths from $x$ to $x$. Composition of morphisms comes from concatenation of paths. The automorphism group of $x$ is the fundamental group $\pi_1(X,x)$, which is a group of symmetries of the based topological space $(X,x)$.
However, if you consider groupoids with many objects all of which are isomorphic to each other, then the different objects correspond to different realizations of the same group. Moreover, each isomorphism between different objects gives rise to an isomorphism between the corresponding realizations of the group.
To make this precise, suppose $\mathcal G$ is a groupoid where all objects are isomorphic. For each object $x$ of $\mathcal G$, let $G_x$ denote the group $\mathrm{Mor}_\mathcal G(x, x)$. A morphism $\phi:x\to y$ defines an isomorphism $G_y\to G_x$ by taking $g\mapsto \phi\circ g\circ \phi^{-1}$.
A nice example of this is the category $\mathrm{FB}_n$ whose objects are finite sets of cardinality $n$ and morphisms are bijections. Then the automorphism group of each object is isomorphic to $S_n$. Isomorphisms between objects determine isomorphisms of their automorphism groups.
Another example is the Fundamental Groupoid of a path connected topological space $X$. Its objects are the points of $X$. The set $\mathrm{Mor}(x,y)$ is the set of homotopy classes of paths from $x$ to $y$. In this groupoid, $G_x$ is the fundamental group $\pi_1(X,x)$ of $X$ based at the point $x$. Different base points result in isomorphic fundamental groups, and isomorphisms are determined by homotopy classes of paths between these points.
Well, I decided to include some additional information.
Definition 1. Let $\mathcal{C}$ and $D$ be categories, $T\colon\mathcal{C}\to\mathcal{D}$ and $S\colon\mathcal{D}\to\mathcal{C}$ be functors. Then the pair $(T,S)$ is called an equivalence iff $S\circ T\cong I_{\mathcal{C}}$ and $T\circ S\cong I_{\mathcal{D}}$. In this case functors $T$ and $S$ are also called equivalences, and categories $\mathcal{C}$ and $\mathcal{D}$ are called equivalent.
It is a basic definition and Tim's answer shows why we need to use equivalence even more frequently than isomorphism. Here's another important definition:
Definition 2. Let $\mathcal{C}$ be a category, $\mathcal{S}$ be a subcategory of $\mathcal{C}$. Then the category $\mathcal{S}$ is called a skeleton of $\mathcal{C}$ iff it is a full subcategory of $\mathcal{C}$ and every object of $\mathcal{C}$ is isomorphic to precisely one object of $\mathcal{S}$.
Note, that if the axiom of choice holds, then every category has a skeleton. See also nLab article. The connection between equivalences of categories and their skeletons is described in the following proposition:
Proposition 1. Let $\mathcal{C}$ and $\mathcal{D}$ be categories, $\mathcal{S}_{\mathcal{C}}$ and $\mathcal{S}_{\mathcal{D}}$ be their skeletons. Then $\mathcal{C}\simeq \mathcal{D}$ iff $\mathcal{S}_{\mathcal{C}}\cong\mathcal{S}_{\mathcal{D}}$.
The proof follows from the fact that every category is equivalent to its skeleton and if two skeletal categories are equivalent, then they are isomorphic. You can also search Mac Lane's "Categories for the working mathematician" for the details.
Thus we can use the notion of skeleton instead of the original definition of equivalence, but sometimes it is not a simplification. As it was mentioned, even an attempt to prove that a category has a skeleton may lead to the set-theoretical difficulties. Tim also gave arguments.
You write: But why go through this whole process of defining this notion of equivalent categories if we can just create a single equivalence classes of objects via the equivalence relation of being isomorphic, and make morphisms defined on the same equivalence class "the same?"
Okay, it could be a good idea if we want to define something like a skeleton. But it isn't, because the straightforward applying this idea leads to wrong definition. Let's try to do this.
Definition 3. Let $\mathcal{C}$ be a category. Then define the graph $\text{Equiv}(\mathcal{C})$ in the following way: $\text{Obj}(\text{Equiv}(C))=\text{Obj}(\mathcal{C})/\cong_{\mathcal{C}}$ and $$\text{hom}_{\text{Equiv}(\mathcal{C})}([a],[b])=(\coprod_{a'\in[a],b'\in[b]}\text{hom}_{\mathcal{C}}(a',b'))/[(f\sim g)\Leftrightarrow(\exists a,b\in \text{Iso}(\mathcal{C})|\quad g\circ a=b\circ f) ].$$
But the graph $\text{Equiv}(\mathcal{C})$ doesn't inherit the composition law from $\mathcal{C}$. The graph $\text{Equiv}(\mathcal{C})$ doesn't even coincide with graph of any skeleton of $\mathcal{C}$ in general case. For example, it may paste two morphisms with the same domain: in the category $\mathbf{Finord}$ we have $\text{end}_{\text{Equiv}(\mathbf{Finord})}([2])=\text{hom}_{\text{Equiv}(\mathbf{Finord})}([2],[2])\cong2$, but $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.
Best Answer
If a group is thought of as a category with just one object, which we might denote *, then an element of the group becomes a morphism from * to itself (so is an automorphism of the object *).
The `category' $\mathbb{Z}_3$ has a single object. The three elements of $\mathbb{Z}_3$ are morphisms from * to *. 0 is the identity morphism on * and $1:*\to *$ is another morphism, and the composition of $1$ with itself gives us $2:*\to *$. NB: * is just an abstract object and is not $\mathbb{Z}_3$ as you tried to write. About the only thing you can know about this object is that it has exactly two endomorphisms other than the identity and they are both invertible, so they are abstract automorphisms. The automorphisms of * form a group isomorphic to $\mathbb{Z}_3$.
I think your final questions are based on a confusion, so they do not quite make sense.
The related question: Confused about the definition of a group as a groupoid with one object. may help.