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If a first-order theory has an infinite model, then for every total order $I$ it has an elementary extension containing distinct order indiscernibles $(a_i{)}_{i\in I}$ whose Skolem hull is a model of the theory. Every order automorphism of $I$ extends uniquely to an automorphism of that hull.

The Ehrenfeucht-Mostowski theorem says that if a first-order theory $T$ has an infinite model, then for every total order $I$ there is a model $M\models T$ generated as the Skolem hull of distinct order indiscernibles $(a_i{)}_{i\in I}$, and every order automorphism of $I$ extends to an automorphism of $M$.

Given an infinite cardinal $\kappa$, let $I=\kappa \times \mathbb{Q}$ with the lexicographic order, viewed as $\kappa$ consecutive copies of the rational order. In each copy independently choose either the identity or a fixed nonidentity order automorphism of $\mathbb{Q}$. These choices give $2^{\kappa }$ distinct order automorphisms of $I$.

Apply the theorem to this order. Distinct order automorphisms act differently on the distinct generators $a_i$, so their extensions give an injection into the automorphism group of a first-order structure $\mathrm{Aut}(M)$. Therefore $|\mathrm{Aut}(M)|\ge 2^{\kappa }$, proving that $T$ has models with arbitrarily large automorphism groups.

Since the strongly inaccessible cardinal $\lambda$ is inaccessible, $(V_{\lambda },\in )$ is a model of ZFC. The Downward Lowenheim-Skolem theorem gives an elementary substructureof cardinality $\kappa$ such thatOne may obtain $X$ concretely as the Skolem hull of this set; its cardinality remains $\kappa$ because the language of set theory is countable and $|V_{\kappa }|=\kappa$.

Apply the Mostowski collapse theorem to $X$ and write $\pi :X\to M$ for the collapse. Then $M$ is a transitive set, $|M|=\kappa$, and $\pi$ fixes $V_{\kappa }$ pointwise. It also fixes $\kappa$, because it fixes every ordinal below $\kappa$. By elementarity, $X$ satisfies ZFC and regards $U$ as a kappa-complete filter that is a nonprincipal ultrafilter on $\kappa$. Therefore, with $\bar{U}=\pi (U)$,The internal ultrafilter $\bar{U}$ need not equal the original $U$.