TITLE: Is Set auto-similar?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.]