A nonempty class of isomorphic groups defines a group The context of this question is from the definition of the sporadic Mathieu group $M_{23}$, which (in one possible definition) is the stabilizer of a point in $M_{24}$, which is a certain subgroup of $S_{24}$ (a permutation group on $24$ points). When I read this I was a bit surprised since they haven't specified which point is to be stabilized, but I assume that the content here is that it doesn't matter, since different stabilizers yield isomorphic subgroups.
Which brings me to the question: is it possible to define a single group given a set of isomorphic groups? Of course in this case we could define $M_{23}$ as the stabilizer of the first point, assuming that the $24$ original points are distinguishable to begin with, for example if they were the numbers $1,\dots,24$ and I could pick out $1$ specifically. But in more general contexts this can be problematic, especially for sets which have no distinguished members, such as the set of all free ultrafilters of $\Bbb N$.
If I have a family $(G_i)_{i\in I}$ of groups for some nonempty index set $I$ such that $G_i\cong G_j$ for all $i,j\in I$, is it possible to define a group $G$ such that $G\cong G_i$ for all $i$? By the observation above, if there is some term $x$ for which $x\in I$ is provable, then it would suffice to take $G=G_x$, but what if there is no such $x$?
 A: Define in specific detail? Of course not. You can choose one from the collection.
But why choose? Why do anything at all? We can revisit the definition of cardinal in $\sf ZF$, as well cardinal arithmetic that goes along with it. Specifically, you might not be able to have a choice of representatives there.
Instead we abstract the notion of cardinal, we assign a certain set, using Scott's trick, to the equivalence class (proper class) of sets under equipollence. We then proceed to calculate addition and multiplication and so on, based on picking an arbitrary representatives from two classes, and showing that we can find a set with the wanted properties defined from those.
Here we can do the same thing. If $\{(G_i,\cdot_i)\mid i\in I\}$ is a set of groups which are pairwise isomorphic then we can say that they define a group $G$ which is abstract. In order to do some calculation in $G$ you pick an arbitrary $G_i$ to do the calculation in, and use the fact that there is an isomorphism between every two, so the result doesn't matter.
In some sense, this group is "pointless", and we only interested in knowing that it is isomorphic to those $G_i$'s, not what its points are. If you have two "pointless groups", $G$ and $H$ their product can be defined as the set $\{G_i\times H_j\mid i\in I,j\in J\}$ (where $\{ H_j\mid j\in J\}$ correspond to $H$, of course). 
And just like in cardinal arithmetic over $\sf ZF$, if you already know that every group can be represented as a "pointless group", then you can just pick two representatives and consider the pointless group representing the product of these two groups. Of course, using Scott's trick you can define canonical pointless groups like that.
