In ZFC, cardinality of set of linear orders over $\omega$ is $2^{\aleph_0}$. By the argument given by here, we can prove (without the choice) the number of linear orders over $\omega$ is at least $2^{\aleph_0}$. In addition, we can prove if $A$ is a set of countable linear order-types then $|A|\ge \aleph_1$ in ZF.
Therefore we can prove $|A|\ge 2^{\aleph_0}$ and $|A|\ge\aleph_1$ in ZF. If we assume the choice then we can prove $|A|\le 2^{\aleph_0}$. However, if we assume the AD (in fact, it is enough that assuming $\aleph_1$ and $2^{\aleph_0}$ are incomparable) then $|A|>2^{\aleph_0}$.
My question is : if we assume the AD (or, every subset of reals is Lebesgue measurable) then $|A|=2^{\aleph_0}+\aleph_1$? If not, there are known results about the cardinality of set of countable linear-order types?
The set of countable linear-order types are defined as follows : Let $C\subset \mathcal{P}( \omega\times \omega)$ be a set of all linear order over $\omega$. Define $\le_1\,\sim\, \le_2$ iff $(\omega,\le_1)$ and $(\omega,\le_2)$ is isomorphic (as linearly ordered set.) Then $\sim$ is a equivalence relation over $C$, and $C/\sim$ is a set of all 'linear order-types' over a countable set.
Nice question!
To get us started, a simple variant of the argument at that link gives us that there is an injection from $\omega_1^{<\omega_1}$ into $C/\sim$: Given such a sequence $(\alpha_\iota\mid \iota<\beta)$, consider the ordered sum $$\underbrace{\alpha_0+(\omega^*+\omega)+\alpha_1+(\omega^*+\omega)+\dots}_\beta,$$ where the underbrace is simply my poor notation to indicate that the sum continues transfinitely to include all $\alpha_\iota$ in the sequence.
The set $\omega_1^{<\omega_1}$ of (well-ordered) countable sequences of countable ordinals is much larger than $\mathfrak c+\omega_1$ (see for instance Woodin's paper on The cardinals below $|[\omega_1]^{<\omega_1}|$).