Isomorphism Classes of Total Orders

Why is this not true? (So they say in mathoverflow).

\begin{Proposition}[Completeness Results]
\label{Completenes Results}
    Let $T_\preceq$ be the set of formulae (i) - (iv) defining Total Orders \ref{Total Order} and consider the classes of structures of $T_\preceq$, $(X, \preceq)$ in the first order signature $\langle \preceq \rangle$ such that $|X| = k$, then note that $T_\preceq \cup \{\vf\}$ there are at maximum eight isomorphism classes such that precisely those three formulae are either theorems or their negation is:
    \begin{itemize}
        \item[($d$)] $\forall_{x, y} \exists_z x \preceq z \preceq y$
        \item[($b^*$)] $\exists_x \forall_y y \preceq x$
        \item[($b_*$)] $\exists_x \forall_y x \preceq y$
    \end{itemize}
\end{Proposition}

I believe that what I wrote is correct but that what they claim is different. What they claim is for a signature $⟨N,⪯⟩$ where $N$ is a name for each of the objects in the domain (i.e. $|N|=k$), which is certainly not the case what I am arguing for.

I work for a prove of this result whose validity I seem to remember from Model Theory (Lecture) in the project Ontic Structuralism of Concepts. The Proposition is in fact almost true, see StackExchange, the result for DLO is also widely known. The proposition is completely true for Elementary Equivalence instead of Isomorphsm, for instance take $(\mathbb{Z},\le )\not\simeq (\mathbb{Z},\le )$, or is it?