Let $\lambda\leq\kappa$ be cardinals, with $\kappa\geq\omega$. I am curious about the number of subsets of $\kappa$ whose order-type (under $\in$) is $\lambda$. Let's denote this by $S(\kappa,\lambda)$; clearly, $S(\kappa,\lambda)\leq\kappa^\lambda$.
CASE 1: $\lambda\leq cf(\kappa)$.
Then for any function $f:\lambda\rightarrow\kappa\setminus\{0\}$, we have a normal sequence $g:\lambda\rightarrow\kappa$ given by $\phi_f(\alpha)=\sum_{\beta<\alpha}f(\beta)$. If $f\neq g$, let $\alpha$ be least such that $f(\alpha)\neq g(\alpha)$. Then $\phi_f(\alpha+1)\neq\phi_g(\alpha+1)$. Therefore, there are exactly $\kappa^\lambda$ normal sequences of length $\lambda$ in $\kappa$; hence $S(\kappa,\lambda)=\kappa^\lambda$.
CASE 2: $\lambda>cf(\kappa)$.
Let's look specifically at $S(\aleph_\omega,\aleph_1)$. First note that $g:\omega_1\rightarrow\aleph_\omega$ cannot be increasing and unbounded; if it were, the function $h:\omega\rightarrow\omega_1$ given by $h(n)=\min\{\alpha<\omega_1:g(\alpha)\geq\aleph_n\}$ would be cofinal, which is impossible. Hence any subset of $\aleph_\omega$ with order-type $\omega_1$ must be bounded; hence it must be a subset of $\aleph_n$ for some $n<\omega$. By CASE 1, $S(\aleph_n,\aleph_1)=\aleph_n^{\aleph_1}=2^{\aleph_1}\aleph_n$ for all $n>1$ (by repeated use of the Hausdorff formula). Hence, $S(\aleph_\omega,\aleph_1)=2^{\aleph_1}\aleph_\omega$. It follows that, $S(\aleph_\omega,\aleph_1)=\aleph_\omega^{\aleph_1}\Leftrightarrow\aleph_\omega\leq 2^{\aleph_1}$. I think that $\aleph_\omega\leq 2^{\aleph_1}$ is not decidable in ZFC, so I guess that $S(\kappa,\lambda)=\kappa^\lambda$ is also not decidable for at least some values of $\kappa$ and $\lambda$, right?
That's as far as I got, so I'm curious to learn what else is known for the CASE 2 case. Are there any general bounds known? Does anyone know how this question relates to any large cardinal axioms? Also, I might have made a mistake in my work, so I'll appreciate if anyone points one out :)