Finitely many homomorphisms to any finite group implies every subgroup of finite index has finitely many automorphic subgroups

Statement
Suppose $$G$$ is a group with finitely many homomorphisms to any finite group. Then, $$G$$ is also a group in which every subgroup of finite index has finitely many automorphic subgroups.

Definitions used
For convenience, we use the following definitions. For more on why these definitions are equivalent to other definitions, see equivalence of definitions of group with finitely many homomorphisms to any finite group and equivalence of definitions of group in which every subgroup of finite index has finitely many automorphic subgroups:

Proof
The proof is immediate with the chosen definitions.

Given: A group $$G$$ with the property that for any natural number, there are only finitely many normal subgroups of $$G$$ of index equal to that natural number. A normal subgroup $$H$$ of $$G$$ of index $$n$$.

To prove: There are only finitely many subgroups $$K$$ of $$G$$ for which there exists an automorphism $$\sigma$$ of $$G$$ satisfying $$\sigma(H) = K$$.

Proof: