 Solution

The class equation implies that a nontrivial finite $p$-group has nontrivial center of a group. Here a center of order $9$ would make $G/Z(G)$ cyclic; a group with cyclic quotient by its center of a group is abelian, contradicting the hypothesis. ThusThe quotient of order nine must be noncyclic for the same reason. For $x\notin Z(G)$, $⟨x,Z(G)⟩$ has order at least nine and lies in the proper centralizer $C_G(x)$. Lagrange's theorem therefore forces $\boxed{|C_G(x)|=9}$. Also $1\ne G^{\prime }\le Z(G)$, so $G^{\prime }=Z(G)$ and all commutators are central of order dividing three.

Classification and splitting. First suppose $a$ has order nine. Then $N=⟨a⟩$ is normal of index three and $Z(G)=⟨a^3⟩$. For $b\notin N$, conjugation on $N$ is nontrivial, since otherwise $a,b$ would generate an abelian group. Its order divides three, so, after replacing $b$ by its inverse if needed, $ba{b}^{-1}=a^4$. Write $b^3=a^{3k}$. Central commutators and $3\mid \left(\genfrac{}{}{0ex}{}{3}{2}\right)$ give $(a^{i}b{)}^3=a^{3i}{b}^3$; choose $i=-k$ modulo three to obtain an element $c=a^{i}b$ of order three outside $N$. This provesThere is a unique resulting isomorphism type, since the two nontrivial actions of order three are interchanged by reversing the complement's generator.

If there is no element of order nine, every nonidentity element has order three. Choose $a,b$ whose images form a basis of $G/Z(G)$ and put $c=[a,b]$. These elements generate $G$, and $c$ generates $Z(G)$. The subgroup $N=⟨a,c⟩\cong C_3^2$ has index three and is normal. The line $⟨b⟩$ meets $N$ trivially and complements it. Conjugation by $b$ fixes $c$ and sends $a$ to $ac$ or $a{c}^{-1}$; choosing the central generator removes this sign. Thus there is one isomorphism type,with the nontrivial unipotent action on ${\mathbb{F}}_3^2$. These presentations are realized by the indicated semidirect products and have order 27. Their exponents are nine and three respectively, so they are not isomorphic. This proves both existence and completeness of the two cases.

Their ordinary character tables are the same, up to relabelling. In either group each of the 24 noncentral elements has conjugacy class of size three, giving eight noncentral classes and three central singleton classes. Each noncentral class is $xZ(G)$, since conjugates of $x$ differ from $x$ by central commutators and its class has size three. There are nine linear characters, arising from $G/G^{\prime }\cong C_3^2$, and eleven irreducible characters altogether. The two remaining degrees satisfy $d_1^2+d_2^2=27-9=18$, so both are three.

Put $Z(G)=⟨z⟩$ and $\omega =e^{2\pi i/3}$. Label the noncentral classes by nonzero $(i,j)\in {\mathbb{F}}_3^2$. For $(r,s)\in {\mathbb{F}}_3^2$ the nine linear characters have valuesAt a noncentral element, column orthogonality gives a squared column norm of nine. The nine linear characters already contribute nine, so both degree-three characters vanish there. By the Schur lemma, a central element has value $3{\omega }^k$ in an irreducible representation of degree three. The central character cannot be trivial, since that would factor the representation through the abelian quotient. Orthogonality of the identity and $z$ columns then forces the two nontrivial central characters to occur once each. The final two rows, on $1,z,z^2$ and the eight other columns, areThis description gives the same entire character table in both cases. The differing element orders are additional power-map data, which an ordinary character table need not determine.