Is Set auto-similar? Let me first define the following notion of auto-similarity (I'm not using the formalism of signature because I'm not really accustomed with it, so let's stay intuitive): we say that a structure $A$ of type $T$ (set, group, topological space, ...) is auto-similar if there exists a strict subset $B$  of $A$ endowed with the same operations as $A$ but restricted to $B$ (or subspace topology for topological space) making $B$ of the same structure type $T$ as $A$, and an isomorphism of structure $T$ between $A$ and $B$.
Ex: $(\mathbb{Z},+)$ is auto-similar because its isomorphic to any $(n\mathbb{Z},+)$ for non zero $n$. The limit ordinal $\omega^\omega$ seen as a monoid whose underlying set is a countable disjoint union of $\mathbb{N}$ with the commutative and associative addition defined as $(i,n) + (i,m) = (i,n+m)$ for all $i$ in $\mathbb{N}$, and $(i,n) + (j,m) =  (j,m) = (j,m) + (i,n)$ iff $i < j$ is auto-similar by any shift sending $(i,m) to (i+j,m)$. The multiplicative monoid $\mathbb{N}$ is also auto-similar by the removal and swap of a finite number of primes and their multiple.
Supposing that our category theory lives in some universe $\mathcal{U}$, we say that a category $\mathcal{C}$ is auto-similar if there exists a subcategory of $\mathcal{C}$ that is equivalent to $\mathcal{C}$. 
Question: is Set auto-similar to a subcategory without any finite set? I'm not really knowledgeable in set theory, but I think one has to suppose GCH here. More precisely, I would like to "shift" all cardinals, sending the first finite set on a countable sets, and $2$ on the first uncountable sets, and so on.
I have thought of the following: send the empty set on the empty set, send $1$ on $\mathcal{P}\mathbb{N}$, send $2$ on $\mathcal{P} \mathcal{P}\mathbb{N}$, send $3$ on $\mathcal{P} \mathcal{P} \mathcal{P}\mathbb{N}$ and so on
I don't know if this construction can be extended to any cardinals, and it doesn't seem easy to find a natural construction for arrows,  but I would guess that it exists supposing GCH.
Maybe someone with good knowledge in set theory have an idea?
 A: This answer is just a straw man that meets the first part of your question about no finite sets, but not the second part about shifting cardinalities: take the subcategory $\cal U$ of $\mathsf{Set}$, comprising all objects of the form $\Bbb{N} \times X$ and all morphisms from $\Bbb{N} \times X \to \Bbb{N} \times Y$ of the form $(i, x) \mapsto (i, f(x))$ where $f$ is a function $X \to Y$. Then $\cal U$ is clearly equivalent to $\mathsf{Set}$ and all its objects are infinite (when viewed from the perspective of $\mathsf{Set}$).
[The following idea doesn't quite work: if you assume your set theory comes equipped with a subcategory of cardinals, $\mathsf{Card}$, say, with exactly one set of each infinite cardinality and exactly one morphism between any two objects, then you can shift cardinalities by taking the objects to be $\kappa(X) \times X$, where $X$ is any set and $\kappa(X)$ is the cardinal equipollent with $\Bbb{N}^{|X|+1}$ and with morphisms from $\kappa(X) \times X \to \kappa(Y) \times Y$ having the form $(i, x) \mapsto (t(j), y)$, where $t$ is the unique morphism in $\mathsf{Card}$ from $\kappa(X) \to \kappa(Y)$. I am not sure how "natural" you would consider such a construction to be. And unfortunately the subcategory $\mathsf{Card}$ we need does not exist: if $\alpha < \beta$ are cardinals, then $1_{\beta} = t \circ s$ where $s : \beta \to \alpha$ and $t : \alpha \to \beta$ is impossible.]
A: I'm not sure I totally understand the question, but I think I can help. 
If $V$ is a model of ZFC, there may be a proper substructure which also models ZFC. In particular, your point about iterating power sets (and then taking limits) leads one to the idea of strongly inaccessible cardinals. Their existence is independent of ZFC because we can use them to build inner models of ZFC. If $\kappa$ is strongly inaccessible, then the set of all sets with 'rank' less than $\kappa$ is an example of such an inner model. 
Alternatively, you can look at other thinks like the constructible universe ($L$) and assume that $V$ is not constructible.
