$\newcommand{\Grp}{\mathrm{Grp}}$Consider the category of groups $\Grp$, and within it we have the solvable groups $S$. Then any group $G$ determines the functor from solvable groups: $$h\_G:=\text{hom}\_{\Grp}(\\_,G):S^{\mathrm{op}}\rightarrow \text{Set}$$


Does this functor determine the group $G$? More concretely, if we have a natural bijection $$\text{hom}\_{\Grp}(A,G)\cong \text{hom}\_{\Grp}(A,G')$$ for all solvable groups $A$, must this be induced by an isomorphism $G\rightarrow G'$?


Are there any circumstances/more restrictive hypotheses where the answer to this is known to be affirmative, for conceptual reasons?


By other circumstance, I mean imposing finiteness conditions, or other "well behavior conditions", or looking at group objects in a more exotic category, etc. I mean conceptual reasons in the sense of being independent of a classification result of all the objects involved, it wouldn't surprise me if this result were true for finite groups, but only verifiable by induction/case checking for the simple groups. Though interesting, I am more interested in any setting where we have a well understood reason for this to hold, or counterexamples/obstacles to its potential truth.



Are groups determined by their morphisms from solvable groups?
Let $G, G'$ be two non-isomorphic Tarski monsters of prime exponent $p$ or two non-isomorphic torsion-free Tarski monsters. Then for every solvable group $A$, $\mathbb{hom}(A,G)\cong \mathbb{hom}(A,G')$.


