Isomorphic to inner automorphism group not implies centerless

Statement
There can exist a group $$G$$ such that the fact about::center of $$G$$ is nontrivial, but $$G$$ is isomorphic to the quotient group $$G/Z(G)$$ (which is also the fact about::inner automorphism group of $$G$$).

Example of a generalized dihedral group
Let $$H$$ be the group of all $$(2^n)^{th}$$ roots of unity among the complex numbers, under multiplication, for all $$n$$. Let $$G$$ be the semidirect product of $$H$$ with a cyclic group of order two acting by the inverse map. In other words, $$G$$ is the uses as intermediate construct::generalized dihedral group corresponding to the abelian group $$H$$. Then, we have:


 * The center $$Z(G)$$ of $$G$$ is a group of order two inside $$H$$, namely, the elements $$\pm 1$$: Clearly, no element outside $$H$$ is in the center, because any element outside $$H$$ acts on $$H$$ by the inverse map under conjugation. This leaves elements inside $$H$$. Any central element of $$H$$ must be fixed by conjugation by elements outside $$H$$, namely, the inverse map. But there are only two elements of $$H$$ that are fixed by the inverse map, namely $$\pm 1$$.
 * The quotient $$G/Z(G)$$ is isomorphic to $$G$$: Note that $$H/Z(G)$$ is isomorphic to $$H$$, via the map sending $$x$$ to $$x^2$$ (the map is well-defined because both elements in the coset of $$x$$, namely $$x$$ and $$-x$$, get sent to the same element: $$x^2$$). Further, the inverse map commutes with this map, so it extends to an isomorphism between $$G/Z(G)$$ and $$G$$.