Abstract groups are fascinating objects on their own, but they are only really useful if you can use them on other structures. To do so, you need the concept of a group action.
Conjugacy classes and centralizers are related to a certain type of group action, a group acting on itself by conjugation - that is, $g\cdot x= g^{-1}xg$. This action gives us a lot of information about the group. The conjugacy classes fit into this picture as the orbits of elements of $G$ under this action, while the centralizers are their stabilizers.
Once you get used to how group actions work, it becomes much easier to see the importance of conjugacy classes and centralizers. For an elementary example, you can tell that $\left|\mathcal{O}_x\right|=[G:C_G(x)]$ for any $x$. It's good to know how many elements a given element is conjugate to, and it's very interesting that this number must divide the order of the group.
For a less elementary example, we can look at how groups act on other sets. A representation of a group $G$ is a homomorphism $G\rightarrow \operatorname{GL}(V)$, where $V$ is a vector space. Usually we are most interested in representations into $\operatorname{GL}_n(\mathbb{C})$. A character is what we call the trace of a representation.
Characters are not necessarily homomorphisms, but, as it turns out, they are class functions, meaning that if $\chi$ is a character, then $\chi(x)=\chi(y)$ whenever $x$ and $y$ are in the same conjugacy class of $G$. So, if we know all the conjugacy classes of the group, we can much more easily compute the characters of the group's representations (and tell a lot of other things about them too). You can see why it would be useful to do this, as symmetries of vector spaces are essential to mathematics in general.
Let $G$ act on itself by conjugation. The orbit stabilizer theorem says
$$
|G\cdot x|=\frac{|G|}{|\operatorname{Stab}(x)|}.
$$
Here the orbit $G\cdot x$ is precisely the conjugacy class of $x$, denote it $C_G(x)$, and the stabilizer $\operatorname{Stab}(x)$ is precisely the centralizer of $x$, denote it $Z_G(x)$.
Let $x\in G$ have a conjugacy class of order $4$, then necessarily its centralizer $Z_G(x)$ has order $3$ by the above. But clearly $Z(G)\subseteq Z_G(x)$, so the center $Z(G)$ either has order $1$ or $3$. If $|Z(G)|=3$, then $x\in Z(G)$, but then it would have trivial conjugacy class as a central element, contrary to $|C_G(x)|=4$. So $Z(G)$ is trivial.
Best Answer
Well, $\langle x\rangle$ is certainly a subgroup of its centralizer $Z(x)$ (which does have order 7) and a group of order $7$ can have only one proper subgroup. If $x=1$, then its class size would be $1$, because the centralizer of $1$ is the entire group.