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.
I believe the comment by James is correct, these groups are precisely the $2$-Engel groups.
Claim: The following statements are equivalent for a group $G$.
Every centralizer in $G$ is a normal subgroup.
Any two conjugate elements in $G$ commute, ie. $x^g x = x x^g$ for all $x, g \in G$.
$G$ is a $2$-Engel group, ie. $[[x,g],g] = 1$ for all $x, g \in G$.
Proof:
1) implies 2): $x \in C_G(x)$, thus $x^g \in C_G(x)$ since $C_G(x)$ is normal.
2) implies 3): $x^g = x[x,g]$ commutes with $x$, thus $[x,g]$ also commutes with $x$.
3) implies 1): If $[[x,g],g] = 1$ for all $g \in G$, then according to Lemma 2.2 in [*], we have $[x, [g,h]] = [[x,g],h]^2$. Therefore $[C_G(x), G] \leq C_G(x)$, which means that $C_G(x)$ is a normal subgroup.
[*] Wolfgang Kappe, Die $A$-Norm einer Gruppe, Illinois J. Math. Volume 5, Issue 2 (1961), 187-197. link
Best Answer
This holds for all elements, and is a consequence of the Orbit-Stabilizer Theorem.
Let $G$ act on its elements by conjugation. Then the orbit of an arbitrary element $g$ is the size of its conjugacy class, and the stabilizer of $g$ under this action is $C_{G}(g)$.