Suppose $G$ is a group. Let's call a total recursive function $F: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ a functional representation if, for all $i \in \mathbb{N}$, the function $n \mapsto F(i, n)$ is a bijection and $\langle \{ n \mapsto F(i, n) \}_{i \in \mathbb{N}} \rangle \cong G$. Let's call $G$ functionally constructive iff it accepts a functional representation and let's call $\langle \{ n \mapsto F(i, n) \}_{i \in \mathbb{N}} \rangle < \operatorname{Sym}(\mathbb{N})$ its functional embedding.
Now, suppose $G$ is a functionally constructive group and $H < \operatorname{Sym}(\mathbb{N})$ is its functional embedding. Let's define a functional representation of an automorphism $\alpha \in \operatorname{Aut}(H)$ as a pair of functional representations of $G$, $F_0$ and $F_1$, such that:
$\langle \{ n \mapsto F_0(i, n) \}_{i \in \mathbb{N}} \rangle = \langle \{ n \mapsto F_1(i, n) \}_{i \in \mathbb{N}} \rangle = H$;
$(\forall i \in \mathbb{N}) \quad \alpha((n \mapsto F_0(i, n))) = (n \mapsto F_1(i, n))$.
Let's call an automorphism of $H$ that has a functional representation functionally constructive. Let's call the set of all functionally constructive automorphisms of $H$ as $\operatorname{Aut}_F(H)$. It is not hard to see, that it is a group.
My question is:
Suppose $G$ is a functionally constructive group and $H_0$ and $H_1$ are its functional embeddings. Is it always true that $\operatorname{Aut}_F(H_0) \cong \operatorname{Aut}_F(H_1)$?
For finite $G$ it is obviously true, because all automorphisms of finite groups are functionally constructive for any their functional embedding. However, I have no idea, what happens when $G$ is infinite...