Hello,
I have the following question: Is it possible for a complete $L\_{\omega\_1,\omega}$ sentence $\phi$ to satisfy (a) the (unique) countable model of $\phi$ has $2^{\aleph\_0}$ many automorphisms and (b) there is a model $N$ of $\phi$, $N$ of size $\aleph\_\omega$ and $N$ has $\le\aleph\_\omega$ many automorhphisms? 


So, $\phi$ will have "many" automorphisms in the countable case and "not very many" automorphisms for a model in power $\aleph\_\omega$. Is this possible and if so, can you reference any examples?



$L_{\omega_1,\omega}$ sentence with many automorphism in $\aleph_0$ and few automorphism in $\aleph_\omega$
It appears that we can already do this in first-order logic, without making any use of the infinitary language.


Namely, the unique countable endless dense linear order $\langle\mathbb{Q},\lt\rangle$ has continuum many order automorphisms, but Shelah has reportedly proved that for every uncountable cardinality $\lambda$, there is a rigid endless dense linear order of cardinality $\lambda$, that is, having a trivial automorphism group. (See the remarks on page 347 of [this article](http://www.ams.org/journals/proc/1978-072-02/S0002-9939-1978-0507336-2/S0002-9939-1978-0507336-2.pdf); perhaps someone can give a better reference.) In particular, there will be an endless dense linear order of size $\aleph\_\omega$ with no nontrivial automorphisms at all.


