The following is supposed to be "clear" according to Kueker, but I could not see why. Can anyone help?


Let $A$ be a countable structure with uncountable many automorphisms. Then for every $\vec{a}\in A^{<\omega}$, $(A,\vec{a})$ has a non-trivial automorphism, i.e. there exists some $f:A\rightarrow A$ such that $f\neq id$ and $f(a)=a$, for all $a\in \vec{a}$. 


Note: The argument is part of a larger proof from [Infinitary Logic: In memoriam Carol Karp](http://www.zentralblatt-math.org/portal/en/zmath/search/?q=an:0316.02018&format=complete) 



Countable structures with uncountable many automorphisms
Let $\vec{a} \in A^{<\omega}$. There are uncountably many automorphisms $f$ and only countably many possible values for $f(\vec{a})$, so there must be two different automorphisms $f\_1$ and $f\_2$ with $f\_1(\vec{a}) = f\_2(\vec{a})$. Then $f\_2^{-1} \circ f\_1$ is a nontrivial automorphism fixing $\vec{a}$.


In fact, there are uncountably many automorphisms moving $\vec{a}$ the same way, so fixing one and composing the others with its inverse gives uncountably many nontrivial automorphisms fixing $\vec{a}$.


