For a topological group $X$ by $Aut(X)$ denote the group of topological isomorphisms $h:X\to X$. If $X$ is compact then the compact-open topology turns $Aut(X)$ into an $\omega$-narrow topological group.
We recall that a topological group $X$ is $\omega$-narrow if for any non-empty open set $U\subset X$ of there exists a countable set $A\subset X$ such that $X=AU$.
Problem. Does the automorphism group $Aut(X)$ of an $\omega$-narrow topological group $X$ admit a topology $\tau$ turning $Aut(X)$ into an $\omega$-narrow topological group such that each compact subset $K\subset(Aut(X),\tau)$ is equicontinuous in the sense that for every neighborhood $U\subset X$ of the unit there exists a neighborhood $V\subset X$ of the unit such that $f(V)\subset U$ for all $f\in K$?
Remark. It seems that for any topological group $X$ the automorphism group $Aut(X)$ admits an $\omega$-narrow group topology: just identify $Aut(X)$ with a subgroup of the homeomorphism group $Homeo(\beta X)$ of the Stone-Cech compactification of $X$. Unfortunatley this topologization does not satisfy the equicontinuity condition.
