How big $|\operatorname{Aut}(M)|$ can be, given $|\partial\operatorname{Aut}(M)|$? My apologies: There were a couple of typos in the original question. Hope I got them all.
Let $\kappa$ be an uncountable cardinal of cofinality $\omega$ and $M$ a model of size $\kappa$. We equip $\kappa^\kappa$ with the product topology and let $\operatorname{Aut}(M)\subset \kappa^\kappa$ denote the topological group of automorphisms of $M$. The closure of $\operatorname{Aut}(M)$ under the product topology is denoted by $\overline{\operatorname{Aut}(M)}$ and $\partial \operatorname{Aut}(M)=\overline{\operatorname{Aut}(M)}\setminus\operatorname{Aut}(M)$ denotes the boundary of $\operatorname{Aut}(M)$.
My question is: If we know that (EDIT)$|\partial\operatorname{Aut}(M)|\ge\kappa^+$(EDIT), can we say anything about the cardinality of $\operatorname{Aut}(M)$, other than (EDIT)$|\operatorname{Aut}(M)|\le|\partial\operatorname{Aut}(M)|$(EDIT)?
Side note: If $f$ is in $\partial\operatorname{Aut}(M)$, then $f$ is 1-1, not onto, and for every formula $\phi$ and every finite $\vec{a}$, $M\models\phi[\vec{a}]$ iff $M\models\phi[f(\vec{a})]$. I.e. $f$ is an elementary embedding.
 A: By Trevor's comment, $\DeclareMathOperator{\Aut}{Aut}|\Aut(M)|\leq|\partial \Aut(M)|$.
From purely topological considerations we get this:
Let $\lambda=\Aut(M)$.
The closure of $\Aut(M)$ cannot have more than $2^{2^\lambda}$ elements.
If $\kappa^+\leq|\partial \Aut(M)|$, then we must have $2^{2^\lambda}\geq\kappa^+$.
So in particular, if GCH or something similar holds below $\kappa$, we have $|\Aut(M)|\geq\kappa$.
A: I was able to prove the following for $\kappa$ of cofinality $\omega$:
Assume $M$ is a model of size $\kappa$, $cf(\kappa)=\omega$ and for all $\alpha<\beta<\kappa^+$, there exist functions $j_{\beta,\alpha}$ in $\DeclareMathOperator{\Aut}{Aut}\overline{\Aut(M)}^{T}\setminus \Aut(M)$, such that for $\alpha<\beta<\gamma<\kappa^+$,
$$(*) j_{\gamma,\beta}\circ j_{\beta,\alpha}=j_{\gamma,\alpha},$$
where $\overline{\Aut(M)}^{T}$ is the closure of $\Aut(M)$ under the product topology in $\kappa^\kappa$.
Then there are at least $\kappa^\omega$ automorphisms of $M$.
The proof is using Infinitary Logic and the assumption that $cf(\kappa)=\omega$ is fundamental.
I do not know of any way to get the above result directly.
(link to arXiv: arXiv:1211.7145).
