A categorical method to, say, determine the cardinality of a group

Every object $c$ in every category satisfies a universal property; it's universal for maps into or out of $c$, or in other words it represents the functors $\text{Hom}(-, c)$ or $\text{Hom}(c, -)$. So rule 5 certainly needs to be clarified. Once we have the ability to name any object we want we can of course just name $\mathbb{Z}$ and hence name the underlying set functor $\text{Hom}(\mathbb{Z}, -)$.

In any case, I think your conditions are much too restrictive. I think a natural example of a categorical method of answering this question would be to describe some categorical property which is satisfied uniquely by $\mathbb{Z}$ (but not necessarily a universal property).

In fact here is such a property: it's a theorem that the free groups are precisely the cogroup objects in $\text{Grp}$. $\mathbb{Z}$ is the unique such group which is indecomposable with respect to coproduct.

A bunch of bits and pieces from a bunch of people:

Let us say we are trying to find the cardinality of the Group $G$. First, we select the group, $\bullet$ (unique up to isomorphism) such that for any other group, there is a unique arrow in and out of $\bullet$. Now, we select a group $\mathbb{Z}$, unique up to isomorphism, such that it only has two idempotent homomorhpisms, and it admits at least two morphisms to any other group besides $\bullet$ (thanks Todd Trimble). Now we find how many morphisms there are from $\mathbb{Z}$ to $G$ (special thanks to Steven Gubkin). This is $|G|$.

Proof

The group $\bullet$ is the trivial group. Now, the group $\mathbb{Z}$ of integers satisfies the properties above via the proof here. Now, for any other group $H$ which satisfies the properties, we know that there is a nontrivial morphism $f : H \rightarrow \mathbb{Z}$. Now the image of $f$ will be a group of integers, and so must be multiples of a given integer $n$ (which won't be zero since $f$ is nontrivial.) We can take a map from this to all the integers, so that we can turn $f$ into a surjection $\bar f$. Now, we make a morphism from $\mathbb{Z}$ to $H$, $i$, such that $i(1) = x$ for some $\bar f(x) = 1$.

Since $\bar f(i(n)) =\bar f(x+x+x+\dotsb)=\bar f(x)+ \bar f(x)+ \bar f(x)+\dotsb=1+1+1+\dotsb=n$, $\bar f \circ i = id_\mathbb{Z}$. This means that $i \circ \bar f$ is idempotent, and since $i(\bar f(x)) = x$, it is not the zero morphism. Since $H$ has only two idempotents (the zero morphism and $\operatorname{id}_H$), and $i \circ \bar f \neq 0$, $i \circ \bar f = id_H$. (Thanks Slade.) Therefore they are inverses. Therefore we can select $\mathbb{Z}$ up to isomorphism.

For each element of $G$, $x$ we make a morphism from $\mathbb{Z}$ to $G$, $h$, such that $h(n) = n * x$. Also, for any morphism $h: \mathbb{Z} \rightarrow G $, $h(n) = n * h(1)$, where $h(1)$ is an element of $G$. Therefore, the morphisms between the integers and $G$ are in one to one correspondence with the elements of $G$. Therefore $|\operatorname{Hom}(\mathbb{Z}, G)| = |G|$.

$\square$

In the category of all groups, the epimorphisms are precisely the surjective mappings according to this encyclopedia of mathematics structures. Therefore, since the epimorphisms are surjective, one can recover the lattice of all normal subgroups of a group simply by taking all the epimorphisms where we regard two epimorphisms $f:A\rightarrow B_{1},g:A\rightarrow B_{2}$ as being equivalent if there is an isomorphism $i:B_{1}\rightarrow B_{2}$ with $if=g$ and we order the epimorphisms with domain $A$ where if $f:A\rightarrow B_{1},g:A\rightarrow B_{2}$, then $f\leq g$ if there is some epimorphism $j:B_{1}\rightarrow B_{2}$ with $jf=g$.

Similarly, the monomorphisms in the category of all groups are precisely the injective mappings, so one can recover the lattice of all subgroups of a group from the category of groups. If $G$ is a group, then the compact elements in the lattice of subgroups are precisely the compact elements in the lattice of subgroups of $G$. It is easy to see that a group is infinite if and only if it has infinitely many finitely generated subgroups: if $G$ is infinite and $G$ contains an element of infinite order $a$, then $\{\langle a^{n}\rangle|n\in\mathbb{N}\}$ are infinitely many finitely generated subgroups. If $G$ has no element of infinite order, then set $a\simeq b$ if $\langle a\rangle=\langle b\rangle$; then each equivalence class in $G$ has finitely many elements, so $G$ has infinitely many equivalence classes and hence infinitely many finitely generated subgroups.

I now claim that if $G$ is infinite, then there are $|G|$ many finitely generated subgroups of $G$. Clearly, there are $|G|$ many finite subsets of $G$, so there can be at most $|G|$ many finitely many subgroups of $G$. If we let $\simeq$ be the same equivalence relation as before where $a\simeq b$ iff $\langle a\rangle=\langle a\rangle$, then each equivalence class can have at most finitely many elements, so there are $|G|$ equivalence classes in $|G|$. Therefore $G$ has at least $|G|$ finitely generated subgroups. We therefore conclude that for infinite groups, the cardinality $|G|$ is equal to the number of compact elements in the lattice of all subgroups of $G$ which can be described in terms of categories.