What does the isomorphism $G / Z(G) \cong \text{Inn}(G)$ mean? $G$ is a group. $Z(G) = \{g \in G: \forall x \in G (gx = xg) \}$ is called the center of $G$. It is easy to verify that $Z(G) \triangleleft G$. $\text{Inn}(G)$ is the inner automorphism group of $G$. As a consequence of the first isomorphism theorem, we have
$$G / Z(G) \cong \text{Inn}(G).$$ 
The homomorphism involved here is defined as $a \in G \mapsto \sigma_{a} \in \text{Inn}(G)$ where $\sigma_{a}$ is a bijection from $G$ to $G$ with $\sigma_{a}(x) = axa^{-1}$. The details can be found here: Factor Group over Center Isomorphic to Inner Automorphism Group.
The isomorphism is too abstract. Could anyone offer some intuitive (or even visual) explanations?
 A: Well, "conjugation by $a$" shuffles the elements of $G$: each $g$ goes to $aga^{-1}$, right? 
There are two possibilities: (i) if $a$ happens to commute with every element of $g$, then $aga^{-1}$ is the same as $gaa^{-1} = g$, so the "shuffling" is just the identity. (ii) If not, then some element $g$ is sent to a different element, and the shuffling is nontrivial. 
So not every element $a$ of $G$ leads to an interesting shuffle: only the ones that fail to commute with at least one other element. So you might think that the inner automorphisms would correspond to $G - Z(G)$. But in fact, if $a$ produces an interesting automorphism, $\phi$, but $b$ produces the identity, then $ab$ will also produce $\phi$. So really the right correspondence is with $G/Z(G)$. 
Does that help at all? 
Let me add a concrete example: Suppose that $G$ is the group of all rotations and dilations of 3-space, where the rotations leave the origin fixed, and by "dilations" I mean maps like $\mathbf v \mapsto c\mathbf v$, where $c$ is a nonzero scalar. Then the subgroup $D$ of dilations is the center of $G$: if you scale up, rotate, and then scale down, it's the same as just rotating. And if you scale up by $a$, then by $b$, then by $1/a$, it's the same as scaling just by $b$. 
And $G/D$ looks just like $SO(3)$ (i.e., the group of origin-fixing rotations of 3-space, known as the "special orthogonal" group -- "orthogonal" because the matrices are orthogonal; "special" because their determinant is $+1$). Why does this correspond to the inner automorphisms of $G$? Because each rotation $R$ corresponds to an automorphism of $G$, namely, the change of basis induced by $R$. 
I don't know if that helps clarify things or not, but it was worth a shot...
A: I would suggest looking at it in the following way.
If we define $Aut(G)$ to be the group of automorphisms of $G$, i.e.
$$
Aut(G) = \{f : G \to G \mid f \text{ is a bijective homomorphism}\}
$$
then there is a map $\phi : G \to Aut(G)$ given by
$$
\phi(g) = \big(h \mapsto ghg^{-1}\big)
$$
that is, $\phi(g)$ is the automorphism given by conjugation by $g$.
Now, it is (hopefully) clear that the image of $G$ in $Aut(G)$ is exactly the set of inner automorphisms, i.e. $Inn(G)$. What is the kernel? Well, the kernel is exactly those $g \in G$ such that $h = ghg^{-1}$ for all $h \in G$. i.e. the kernel is $Z(G)$!
Thus it follows that $im(\phi) = G / \ker\phi$ (this is a restatement of the first isomorphism theorem), i.e. $Inn(G) = G / Z(G)$.
