Solution set condition and the forgetful functor $U:\textbf{Gr}\rightarrow\textbf{Set}$ Let $U:\textbf{Gr}\rightarrow\textbf{Set}$ be the forgetful functor from the category of groups to the category of sets.
Let $X\in\textbf{Set}$. I want to construct the solution set $S_X\subset\text{Ob}(\textbf{Gr})$ associated to $X$. I propose the following correspondence:
$$S_X:=\left\{G\in\textbf{Gr}\ |\ (\exists i)(X\xrightarrow{i} U(G)\land\text{"$i(X)$ generates $G$"})\right\}.$$
The trouble is that I am not sure how to prove that the class $S_X$ is a set. In his answer to the question Adjoint Functor Theorem, Martin says,

The class of these groups is essentially small, since $U(G)$ admits a
surjection from $\coprod_n(X\times\mathbf{N})^n$.

He uses the same argument in his answer to the question Existence proof of the tensor product using the Adjoint functor theorem. to conclude once more that the solution class is a set; here, he says

Then $\#|A'|\leq\aleph_0\cdot\#|M|\cdot\#|N|$. Hence, up to
isomorphism, there is only a set of such $A's.$

What sort of an argument is this? I am not clear about why showing that $|U(G)|\leq|\coprod_n(X\times\mathbf{N})^n|$, for each $G\in S_X$, allows us to conclude that $S_X$ is a set. Can someone please flesh this out for me?
 A: If you are just interested in finding some solution set, then there is an easier answer. Arguably this is cheating though, because it uses the description of the left adjoint $F$ of the forgetful functor $U: \mathbf{Gr} \to \mathbf{Set}$.
That being said, let's fix some set $X$. Then I claim that $\{F(X)\}$, where $F$ is the free group on $X$, is a solution set. Indeed, let $f: X \to U(G)$ be any function of sets. Then because $F(X)$ is the free group on $X$ we can extend $f$ to a group homomorphism $\bar{f}: F(X) \to G$ such that $\bar{f} i = f$, where $i: X \to F(X)$ is the obvious injection of the generators.

As I said, this is cheating because it uses a description of the left adjoint of $U$. Indeed, the free group functor $F: \mathbf{Set} \to \mathbf{Gr}$ is the left adjoint of $U$. So the above is just a specific instance of the following fact.

Let $L: \mathcal{C} \rightleftarrows \mathcal{D}: R$ and let $C$ be an object in $\mathcal{C}$. Then $\{ L(C) \}$ is a solution set for $R$, associated to $C$.

Proving this would be a nice exercise. Hint: use the unit $\eta_C: C \to RL(C)$.

The class $S_X$ you proposed is of course going to capture the free group as well, so it will definitely contain enough. As already mentioned in the comments, $S_X$ is technically a proper class. However, this is no issue, because it will be essentially small. That means that there is a set $S_X'$ such that for every group $G \in S_X$ there is $G' \in S_X'$ with $G' \cong G$. So technically $S_X'$ would be our solution set, but we might as well work with $S_X$ and suppress the isomorphisms in our notation (so this is purely for convenience).
To prove that this class is essentially small, you quote another answer where they give a cardinality argument. Basically what they say is that there is some cardinal $\kappa$ such that for all $G \in S_X$ we have $|U(G)| \leq \kappa$. If we fix a set $Y$ then there is only a set of group structures on $Y$. So for every cardinal $\lambda \leq \kappa$ there is, up to isomorphism, only a set of group structures where the underlying set has cardinality $\lambda$. Hence there is, up to isomorphism, only a set of groups where the underlying set has cardinality $\leq \kappa$.
In both answers you quoted it is explained how this cardinality bound is found (although I think in the first one it should be $\mathbb{Z}$ in place of $\mathbb{N}$, but this is of no real consequence).
