Ordinal-universal linear order on $\kappa$ elements The starting point of this question is the observation that if $\lambda$ is a countable ordinal, then there is an order-embedding $e:\lambda \hookrightarrow \mathbb{Q}$.
Given an infinite cardinal $\kappa$, is there a linearly ordered set $L_\kappa$ of cardinality $\kappa$ such that for every ordinal $\alpha$ with $|\alpha|=\kappa$ there is an embedding $e:\alpha \hookrightarrow L_\kappa$?
 A: Let $\lambda$ be an infinite cardinal. Give $[\lambda]^{<\omega}$ the linear ordering $\leq$ where we set $A>B$ if there is some $\alpha\in B$ where $\alpha\cap A=\alpha\cap B$ and $\alpha\not\in A$.
We shall show using what is essentially transfinite induction that every ordinal less than $\lambda^+$ embeds into $[\lambda]^{<\omega}$. Suppose that $\alpha$ is the least ordinal where $\alpha$ does not embed into $\lambda^+$. Clearly $\alpha>0$. If $\alpha$ is a successor ordinal, then $\alpha=\beta+1$ for some $\beta$ and there is some embedding $f:\beta\rightarrow[\lambda]^\omega$. Now, define $g:\alpha\rightarrow[\lambda]^\omega$ by letting
$g(\delta)=\{0\}\cup\{1+\sigma\mid \sigma\in f(\delta)\}$ whenever $\delta<\alpha$ and by letting $g(\beta)=\emptyset$. Then $g$ is an order preserving embedding, so $\alpha$ cannot embed into $\lambda^+$. Now suppose that $\alpha$ is a limit ordinal. Then let $(x_\beta)_{\beta<\gamma}$ be a cofinal continuous increasing sequence in $\alpha$ with $x_0=0$ where $\gamma$ is a cardinal with $\gamma\leq\lambda$. Then for each $\beta<\gamma$, let
$f_\beta:[x_\beta,x_{\beta+1})\rightarrow[\lambda]^{<\omega}$ be an embedding. Then define an embedding $f:\alpha\rightarrow[\lambda]^{<\omega}$ by letting
$f(x)=\{\beta\}\cup\{\beta+1+\delta\mid \delta\in f_\beta(x)\}$ whenever $x\in[x_\beta,x_{\beta+1})$.
Observe that $[\lambda]^n$ is a well-ordered of order type $\lambda^n$ whenever $n\in\omega$. Therefore $[\lambda]^{<\omega}$ is the union of countably many well-ordered sets. Every totally ordered set $X$ which is the union of countably many well-ordered sets has no descending sequence of length $\omega_1$. Therefore, $[\lambda]^{<\omega}$ has no descending sequence of length $\omega_1$.
A: Let me mention one easy observation, which is that the order $2^{<\kappa}$ is universal for all linear orders of size $\kappa$. View the order as $\{-1,1\}^{<\kappa}$, where strings are ordered lexically, so that at the first difference, if the digit is $-1$ it is lower, and if $1$ it is higher. This is the same as the Surreal number line, generating for birthdays up to $\kappa$.
It often happens in set theory that $2^{<\kappa}$ has size $\kappa$, in which case this order is a linear order of size $\kappa$ that is universal for all linear orders of size $\kappa$. For example, this is true for $\kappa=\omega$, and it is true for any $\kappa$ under the generalized continuum hypothesis.
Meanwhile, it is consistent with ZFC that $2^{<\kappa}$ is strictly bigger than $\kappa$, and in these cases this order is larger than desired. There are quite subtle set-theoretic issues arising in these cases, discussed in the references mentioned at the other question, mentioned in the comments.
Lastly, you had asked for universality with respect to the ordinals of size $\kappa$, rather than all linear orders of size $\kappa$. I find it interesting to wonder whether these are the same---if a linear order of size $\kappa$ is universal for ordinals up to $\kappa^+$, must it be universal for all linear orders of size $\kappa$?
