**Definition (1):** ‎An ‎‎$‎‎‎\mathcal{L}‎$ -‎ ‎structure ‎‎$‎‎‎\mathcal{M}‎$ ‎called ‎"‎‎rigid" ‎iff ‎‎there ‎is ‎no ‎non-trivial automorphism on ‎$‎\mathcal{M}‎$.‎
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**Definition (2):** ‎An ‎‎$‎‎‎\mathcal{L}‎$ -‎ ‎structure ‎‎$‎‎‎\mathcal{M}‎$ ‎called ‎"non-‎‎rigid" ‎iff ‎it ‎is ‎not a‎ ‎rigid ‎model.‎


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‎**Definition (3):**‎‎ ‎An ‎‎$‎‎‎\mathcal{L}‎$ -‎ ‎structure ‎‎$‎‎‎\mathcal{M}‎$ ‎called ‎"flexible" ‎iff ‎$‎‎‎\mathcal{M}‎$ ‎has ‎the ‎"maximum" ‎number ‎of ‎automorphisms.(For structures with a set as their domain of discourse this means ‎$‎‎|Aut(‎\mathcal{M}‎)|=2^{|‎\mathcal{M}‎|}$ and for the structures with a proper class domain we "define" this notion to be ‎$|Aut(‎\mathcal{M}‎)|=|Ord|‎$ ‎which is an abbreviation for "there are class-many automorphisms"‎‎‎)


**Definition (4):** By replacing the notion of "automorphism" with "self elementary embedding" one can define the "pseudo rigid", "pseudo non-rigid" and "pseudo flexible" structures in the similar way on definitions (1), (2) and (3).
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There ‎are ‎many ‎(set and proper class) rigid ‎models ‎in set theory and ‎model ‎theory (like ‎$‎‎‎\mathbb{R}‎$ as an ordered field or ‎$‎V‎$ ‎as an ‎$‎‎\in$‎-model of ‎$‎ZFC‎$‎‎‎). ‎But ‎even ‎there ‎are ‎many ‎"origamic" model construction methods ‎which ‎can ‎change ‎the ‎nature of a model ‎"softely". Ultrapower is ‎one ‎of ‎these ‎operators. ‎Intuitionally‎‎ ‎the ‎possibility ‎of occuring a non-trivial ‎self‎-similarity in a rigid model increases with corrugating it by larger ultrapowers. Now there are some questions here: 
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**Question (1)**: ‎Let ‎‎$‎‎\mathcal{M}‎$ ‎be a ‎set-kind ‎rigid ‎‎$‎‎\mathcal{L}‎$ -‎ ‎structure. ‎Are ‎there a‎n ‎‎index set $I‎$‎ ‎and a‎n ‎‎ultrafilter ‎‎$‎‎‎\mathcal{F}‎$ ‎on ‎it ‎such ‎that ‎‎$‎‎\Pi\_{‎\mathcal{F}‎}‎‎‎\mathcal{M}‎$ ‎be a ‎"‎‎non-rigid" ‎model (‎$‎‎|Aut(‎‎\Pi\_{‎\mathcal{F}‎}‎‎‎\mathcal{M})|>1$‎)?‎


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**Question (2)**: ‎Let ‎‎$‎‎\mathcal{M}‎$ ‎be a ‎set-kind ‎rigid ‎‎$‎‎\mathcal{L}‎$ -‎ ‎structure. ‎Are ‎there a‎n ‎‎index set $I‎$‎ ‎and a‎n ‎‎ultrafilter ‎‎$‎‎‎\mathcal{F}‎$ ‎on ‎it ‎such ‎that ‎‎$‎‎\Pi\_{‎\mathcal{F}‎}‎‎‎\mathcal{M}‎$ ‎be a ‎"flexible" ‎model (‎$‎‎|Aut(‎‎\Pi\_{‎\mathcal{F}‎}‎‎‎\mathcal{M})|=2^{|‎‎\Pi\_{‎\mathcal{F}‎}‎‎‎\mathcal{M}|}$‎)?‎
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‎**Question (3)**: What are the answers of above questions for an arbitary rigid proper class model (like ‎$‎‎‎\langle ‎‎V,\in\rangle‎‎$‎)? ‎
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**Remark (1):** ‎The notion of the ultrapower for a‎ ‎proper ‎class ‎model can be ‎defined ‎by ‎"Scott's trick" ‎to ‎produce a‎nother ‎proper ‎class ‎model.‎ 


Even there are some questions about possible resolving of "pseudo rigidity" problem of $\langle V,\in \rangle$ by ultraproducts:


**Question (4):** Are ‎there‎ ‎large ‎cardinal ‎axioms like ‎‎$‎A‎$ and ‎$‎B‎$‎ ‎such ‎that:‎
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**(a)** ‎‎$‎‎ZFC+A‎\Longrightarrow ‎‎\exists ‎M‎‎~‎\exists ‎j:‎\langle ‎V,\in ‎\rangle ‎\longrightarrow‎‎ ‎‎\langle ‎M,\in ‎\rangle‎‎‎‎‎‎~~;~~‎\langle ‎M,\in ‎\rangle~is~a~"pseudo~non-rigid"~inner~model~of~ZFC‎$‎‎
\‎$‎~~‎\wedge‎~~j~is~a~non-trivial~elementary~embedding‎‎‎$‎


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**(b)** ‎‎$‎‎ZFC+B‎\Longrightarrow ‎‎\exists ‎M‎‎~‎\exists ‎j:‎\langle ‎V,\in ‎\rangle ‎\longrightarrow‎‎ ‎‎\langle ‎M,\in ‎\rangle‎‎‎‎‎‎~~;~~‎\langle ‎M,\in ‎\rangle~is~a~"pseudo~flexible"~inner~model~of~ZFC‎$‎‎
\‎$‎~~‎\wedge‎~~j~is~a~non-trivial~elementary~embedding‎‎‎$‎‎‎‎‎‎‎
‎ ‎‎
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Can we flex the rigid models by enough power?
$\textbf{Question 1}$ For question $1$, the answer is affirmative.


Suppose that $\mathcal{M}$ is an infinite structure over some language $\mathcal{L}$. Then take a sufficiently large ultrapower $\mathcal{M}^{\mathcal{U}}$ (if necessary) such that there exists distinct points $a,b\in\mathcal{M}^{\mathcal{U}}$ such that the expanded structure $(\mathcal{M}^{\mathcal{U}},a)$ is elementarily equivalent to $(\mathcal{M}^{\mathcal{U}},b)$. Then by the Shelah-Keisler isomorphism theorem [1](http://link.springer.com/article/10.1007/BF02771574), there is some ultrafilter $\mathcal{V}$ and an isomorphism $\phi:(\mathcal{M}^{\mathcal{U}},a)^{\mathcal{V}}\rightarrow (\mathcal{M}^{\mathcal{U}},b)^{\mathcal{V}}$. In particular, $\phi$ is an isomorphism between the reducts $\phi:(\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}\rightarrow(\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}$ that is not the identity map. Furthermore, the composition of ultrapowers is an ultrapower in the sense that there is a unique up to Rudin-Keisler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that for each structure $\mathcal{A}$, we have $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ be isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$. We therefore conclude that there is a non-trivial isomorphism $\phi':\mathcal{M}^{\mathcal{U}\cdot\mathcal{V}}\rightarrow\mathcal{M}^{\mathcal{U}\cdot\mathcal{V}}$.


$\textbf{Question 2}$ Assuming GCH (or weaker assumptions) the answer is also affirmative. 
If GCH holds, then every structure $\mathcal{A}$ has a saturated ultrapower $\mathcal{A}^{\mathcal{U}}$ of sufficiently large cardinality. On the other hand, if $\mathcal{B}$ is a saturated structure over a language $\mathcal{L}$ and $|\mathcal{B}|>|\mathcal{L}|$, then $|Aut(\mathcal{B})|=2^{|\mathcal{B}|}$. In particular, $|Aut(\mathcal{A}^{\mathcal{U}})|=2^{|\mathcal{A}^{\mathcal{U})}|}$ for some ultrafilter $\mathcal{U}$.
In fact, the paper [2](http://arxiv.org/pdf/math/9304201.pdf) shows that for every saturated structure $\mathcal{B}$ over a language $\mathcal{L}$ with $|\mathcal{B}|>|\mathcal{L}|$, there is a free subgroup $\mathcal{C}$ of $Aut(\mathcal{B})$ generated by $2^{|\mathcal{B}|}$ elements where $\mathcal{C}$ is dense in $Aut(\mathcal{B})$. Here dense is used in a topological sense. We have $Aut(\mathcal{B})\subseteq\mathcal{B}^{\mathcal{B}}$ and we give $\mathcal{B}^{\mathcal{B}}$ the topology of pointwise convergence and $Aut(\mathcal{B})$ the subspace topology.


$\textbf{Question 3}$ This question probably could be answered better by someone else. Nevertheless, I will explain what I know about this matter.


Ultrafilters possess a certain amount of rigidity in the following sense. For each ultrafilter $\mathcal{U}$ there is a first order structure $\mathcal{A}$ such that the automorphism group $\mathcal{A}^{\mathcal{U}}$ is trivial. In particular, there is a sufficiently complicated proper class structure $\mathcal{A}$ such that for each set sized ultrafilter $\mathcal{U}$, the structure $\mathcal{A}^{\mathcal{U}}$ is rigid. Unfortunately, the structure $\mathcal{A}$ that I am thinking of contains a proper class of constants (or relation symbols) unlike $(V,\in)$.


On the other hand, this issue of rigidity can be bypassed by considering a the direct limit of a sequence (or a net) of ultrapowers instead of a single ultrapower. More precisely, there exists a sequence of ultrafilters $(\mathcal{U}\_{n})\_{n}$ such that for each first order structure $\mathcal{A}$, the automorphism group of the direct limit of ultrapowers $^{\lim}\_{n\rightarrow\infty}\mathcal{A}^{\mathcal{U}\_{0}\cdots\mathcal{U}\_{n}}$ is non-trivial.


[1](http://link.springer.com/article/10.1007/BF02771574) Shelah, Saharon
[Every two elementarily equivalent models have isomorphic ultrapowers.](http://link.springer.com/article/10.1007/BF02771574) 
Israel J. Math. 10 (1971), 224–233. 


[2](http://arxiv.org/pdf/math/9304201.pdf) Melles, Garvin(IL-HEBR); Shelah, Saharon(IL-HEBR)
[Aut(M) has a large dense free subgroup for saturated M](http://arxiv.org/pdf/math/9304201.pdf) . (English summary) 
Bull. London Math. Soc. 26 (1994), no. 4, 339–344. 


