= Solution
The maximum is $2^{\aleph_1}$. We construct that many <aleph-one-like linear orders>, whose proper initial segments are <countable>, and prove that they are nonisomorphic. As usual the rational-type requirement concerns nonempty proper initial segments; the empty segment occurs in every order.
Let $D$ be the <club set> of nonzero <limit ordinals> below $\omega_1$. For $S\subseteq D$, form the <order sum>
$$
L_S=\sum_{\alpha<\omega_1}B_\alpha,\qquad
B_\alpha=
\begin{cases}
1+\mathbb Q,&\alpha\in S,\\
\mathbb Q,&\alpha\notin S.
\end{cases}
$$
Here $1+\mathbb Q$ means a new least point followed by a copy of the <rational numbers>; distinct blocks are disjoint, and all points of an earlier block precede every point of a later one. Every block is <countable>, dense and has no greatest point. Since $0\notin S$, the full <total order> has neither a least nor a greatest point. It has size $\aleph_1$.
The sum is dense: within a block use density, and for points in different blocks choose a larger point in the earlier block. If $I$ is a proper initial segment and $y\notin I$ lies in block $\alpha$, then $I$ is contained in the union of the blocks through $\alpha$. There are only countably many such blocks, each <countable>. Thus $I$ is <countable>, rather than an <uncountable> interval in an aleph-one-dense order.
Every nonempty $I$ is dense and has no least point. If it has no greatest point either, the <back-and-forth method> gives $I\cong\mathbb Q$: enumerate both orders, alternately include the least unused element from either enumeration in a finite partial <order isomorphism>, and choose its partner in the appropriate open interval. Density and absence of endpoints always supply a new partner. The union is a bijective order isomorphism. If $I$ has a greatest point, remove it, apply the same argument, and reattach it. Hence every nonempty proper initial segment is isomorphic to $\mathbb Q$ or $\mathbb Q+1$.
\b[The stationary information is recorded at the limit cuts.] Write
$$
F^S_\alpha=\bigcup_{\beta<\alpha}B_\beta.
$$
This is a continuous increasing <kappa-filtration> by <countable> initial segments. At a limit $\delta$, the complement $L_S\setminus F^S_\delta$ has a least point exactly when $\delta\in S$.
Suppose $f:L_S\to L_T$ is an <order isomorphism>. Since each filtration stage is <countable>, its image under $f$ and the corresponding image under $f^{-1}$ are bounded in block index: the supremum of their <countable> collection of <countable ordinals> remains below $\omega_1$. Choose $g(\alpha)<\omega_1$ so that
$$
f[F^S_\alpha]\subseteq F^T_{g(\alpha)},\qquad
f^{-1}[F^T_\alpha]\subseteq F^S_{g(\alpha)}.
$$
The set
$$
C=\{\delta\in D:(\forall\alpha<\delta)\ g(\alpha)<\delta\}
$$
is a <club set>. To see unboundedness, start above any given ordinal and recursively choose $\gamma_{n+1}>\gamma_n$ larger than every $g(\alpha)$ for $\alpha\le\gamma_n$. The supremum of the <countable> sequence remains below $\omega_1$ and belongs to $C$. Closure follows because below a limit of members of $C$, each fixed $\alpha$ lies below one of those members.
For $\delta\in C$, continuity of the filtrations gives both inclusions needed for
$$
f[F^S_\delta]=F^T_\delta.
$$
The isomorphism therefore preserves whether the complementary final segment has a least point. Consequently
$$
\delta\in S\quad\Longleftrightarrow\quad\delta\in T
\qquad(\delta\in C).
$$
This <club agreement of countable filtrations> proves the invariant: if $S\mathbin{\triangle}T$ is stationary, the orders cannot be isomorphic.
We also construct enough pairwise different stationary encodings. For each $\beta<\omega_1$, fix an <injection> $e_\beta:\beta\to\omega$, and set
$$
A_{\xi,n}=\{\beta\in D:\xi<\beta,\ e_\beta(\xi)=n\}.
$$
For fixed $\xi$, these sets partition the stationary tail $D\setminus(\xi+1)$. At least one $A_{\xi,n}$ is stationary: otherwise choose a club avoiding each one and intersect the countably many clubs to obtain a club avoiding the entire tail. The countable-intersection fact follows directly: above any starting point choose an increasing sequence, visiting each club infinitely often; its <countable> supremum belongs to every club by closure. For each $\xi$ let $n(\xi)$ be the least such $n$. One value $n_*$ occurs for uncountably many $\xi$, since a <countable> union of <countable> sets is <countable>. For that fixed $n_*$ the cells are pairwise disjoint, by injectivity of $e_\beta$. Enumerating those cells gives
$$
(S_\eta:\eta<\omega_1),
$$
a family of pairwise disjoint stationary subsets of $D$. This is the explicit <Ulam matrix on omega-one> proof of <disjoint stationary subsets of omega-one>.
For $A\subseteq\omega_1$ put $S(A)=\bigcup_{\eta\in A}S_\eta$. If $A\ne A'$, their symmetric difference contains an entire $S_\eta$ and is stationary. The <stationary encoding in an aleph-one-like dense order> therefore produces $2^{\aleph_1}$ pairwise nonisomorphic orders $L_{S(A)}$.
Finally, on a fixed carrier of size $\aleph_1$, every total order is a subset of its Cartesian square. That square has size $\aleph_1$, so there are at most $2^{\aleph_1}$ possible relations. Thus
$$
\boxed{2^{\aleph_1}\text{ pairwise nonisomorphic orders, the maximum possible}.}
$$
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