Endomorphisms of cardinals

Question

Let $\kappa$ be a cardinal. Viewing $\kappa$ as an ordered set, let $\operatorname{End}(\kappa)$ be the set of endomorphisms of $\kappa$: $$ \operatorname{End}(\kappa):=\{\ f:\kappa\to\kappa\ |\ (\forall\ \alpha,\beta\in\kappa)\ \alpha\le\beta\implies f(\alpha)\le f(\beta)\}, $$ and let $\kappa'$ be the cardinality of $\operatorname{End}(\kappa)$.

Do we necessarily have $\kappa'>\kappa$?

Note that we have $\kappa'\ge\kappa$ for all $\kappa$ (because of the constant endomorphisms), and $\kappa'>\kappa$ if $\kappa\le\aleph_0$ (because if the identity endomorphism if $\kappa<\aleph_0$, and by a straightforward argument if $\kappa=\aleph_0$), so that a counterexample would necessarily be larger than $\aleph_0$.

A counterexample would also answer this question.

A slightly more ambitious variant would be: Compute $\kappa'$ for each $\kappa$. We have $$ \kappa'=\binom{2\kappa-1}{\kappa-1} $$ for $\kappa$ finite, $(\aleph_0)'=2^{\aleph_0}$, and $\kappa'\le2^\kappa$ for all $\kappa$.

Answer 1

Recall $\{A\subseteq\kappa\mid |A|=\kappa\}$ has size $2^\kappa$ (simply because we can split $\kappa$ into two parts of size $\kappa$, and then just add subsets from the one part to the second part).

Now, for every $A\subseteq\kappa$, if $|A|=\kappa$ then there is a unique isomorphism between $A$ and $\kappa$. Suppose that $f\colon\kappa\to A$ is such isomorphism, then it is in $\operatorname{End}(\kappa)$.

Therefore there are at least $2^\kappa$ such endomorphisms. On the other hand, there cannot be more, since for any infinite $\kappa$, $\kappa^\kappa=2^\kappa$.

Answer 2

Let $\mathrm{Lim}$ be the class of limit ordinals. Then $|\mathrm{Lim}\cap \kappa|=\kappa$ for every uncountable cardinal $\kappa$, let $\kappa_L$ denote $\mathrm{Lim}\cap\kappa$.

Let $A$ be a subset of $\kappa_L$ and let $f_A\colon\kappa\to\kappa$ be the function defined by $f(\gamma+1)=\gamma$ if $\gamma\in A$ and $f(\gamma)=\gamma$ otherwise. It's clear that $f_A$ is an endomorphism of $\kappa$ and that $A\neq A'$ implies $f_A\neq f_{A'}$, hence there are at least $|\mathcal{P}(\kappa_L)|=|\mathcal{P}(\kappa)|>\kappa$ such endomorphisms.