A capable group is a group that is isomorphic to $G/Z(G)$ for some group $G$.
So, is every member of the OEIS sequence A216594 (consisting of those $n$ for which there is a non-nilpotent group, but no centerless group, of order $n$) the order of a capable group?
Every even number $2n$ (for $n>1$) is the order of a capable group. Indeed, the dihedral group of order $2n$ is isomorphic to the central quotient of the dihedral group of order $4n$ as pointed out at Non-cyclic numbers that are not the order of any capable group.
So, only odd numbers need to be considered. The first odd member of A216594 is $63$. Since $63$ is not a powerful number (it is divisible by the prime $7$ but not by $7^2=49$), there is no capable abelian group of order $63$ by the known characterization of finite capable abelian groups (isomorphic to a direct sum of cyclic groups, with the order of each one dividing the order of the next one and the last two being isomorphic, or equivalently, having the same order).
So, to try to find a capable group of order $63$, one must consider nonabelian groups. If $G$ is the nonabelian group of order $21$ (given as the semidirect product $\mathbb{Z}_7 \rtimes \mathbb{Z}_3$ with $(a, b)(c, d)=(a+2^bc, b+d)$), then one could consider the direct product $G \times \mathbb{Z}_3$. The other nonabelian group of order $63$ is given by the semidirect product $\mathbb{Z}_7 \rtimes \mathbb{Z}_9$ (again with $(a, b)(c, d)=(a+2^bc, b+d)$). If neither of those two nonabelian groups of order $63$ turns out to be capable, then the answer to the question must be "no".
Otherwise, one could consider the next odd member of A216594, namely $105$, which is also the first squarefree member. The only nonabelian group of order $105$ is given by the direct product of the nonabelian group of order $21$ with the cyclic group of order $5$. If that group of order $105$ turns out to not be capable, then the answer to the question must again be "no".
If neither $63$ nor $105$ succeeds in finding a member of A216594 that is not the order of any capable group, then one could consider $117$ (the next odd member) or $165$ (the next odd squarefree member). But $117$ and $165$ have the same "nilpotent number relationships" (of the form $p^k \equiv 1 \pmod q$) among their prime factors and the same prime signature as $63$ and $105$ respectively, so I do not think that $117$ or $165$ would succeed if $63$ and $105$ did not succeed. So, the next order to consider would be $189$.
Best Answer
The group of order $63$ with elementary abelian Sylow $3$-subgroup is capable - it is isomorphic to $G/Z(G)$ with $|Z(G)|=3$.
However the nonabelian group of of order $105$ is not capable, because it has the form $H \times C_p$ where $|H|$ is coprime to the prime $p$.
Suppose such a group was isomorphic to $G/Z(G)$. Then the inverse image of the $C_p$ factor of $G/Z(G)$ would be abelian and normal in $G$, and $G$ would have a normal Sylow $p$-subgroup $P$. A complement of $P$ is $G$ would centralize both $P \cap Z(G)$ and $P/P \cap Z(G)$ and hence centralize $P$, so $P$ would be a direct factor of $G$ and hence in $Z(G)$, contradicting the structure of $G/Z(G)$.