I am not sure the importance of A being unary here. If anyone can explain if the following is the right idea.
So let A be some unary algebra, to be specific not necessarily a mono-unary algebra. I mean that A may have more than one operation but all operations are unary. so if $\sigma$ is an automorphism on A then $\forall a \in A$ we have: $\sigma(a) = b, b \in A$ and for all $f$ such that it is an operation on $A$ $\sigma(f(a)) = f(\sigma(a))$ additionally, $\sigma$ is one-to-one and onto. then the group $\langle Aut(A), \circ, id \rangle = \{ \sigma : \sigma $ is an automorphism on $A \}$ where $\circ $ is composition and $id$ is the identity automorphism.
So if $G$ is some group isomorphic to $Aut(A)$ then we have a map $X$ where
$X : G \to Aut(A)$ such that it is homomorphic and bijective. It seems like something such as left translation (left multiplication) is the key. i.e. that the unary operations on A are analogous to left translation (or similar) in G.
What you've written so far shows that you're thinking about the problem the right way, but you'll have to do more work to come up with the construction of $A$ and the proof that it works. You're absolutely right that left translation is important. See below for a full solution to the problem.
Let $G$ be a group with identity $e$. Consider the unary signature $\{f_g\mid g\in G\}$. Let $A = G$, and interpret the unary operations as right translations: for $a\in A$, $f_g(a) = ag$.
There is a map $\varphi\colon G\to \mathrm{Aut}(A)$, given by left translation: For $h\in G$, $\varphi(h) = \sigma_h$, where $\sigma_h(a) = ha$. Let's check that $\sigma_h$ is an automorphism of $A$: It is a homomorphism, since for all $g\in G$, $\sigma_h(f_g(a)) = hag = f_g(\sigma_h(a))$, and it is invertible with inverse $\sigma_{h^{-1}}$.
Now $\varphi$ is a homomorphism, since $\varphi(gh)(a) = \sigma_{gh}(a) = gha = (\sigma_g\circ \sigma_h)(a) = (\varphi(g)\circ \varphi(h))(a)$. It is injective, since if $g\neq h$, then $\varphi(g)(e) = ge = g \neq h = he = \varphi(h)(e)$. For surjectivity, let $\sigma\in \text{Aut}(A)$. I claim that $\sigma = \varphi(\sigma(e))$. Indeed, for any $a\in A$, $$\sigma(a) = \sigma(ea) = \sigma(f_a(e)) = f_a(\sigma(e)) = \sigma(e)a = \varphi(\sigma(e))(a).$$