Let $G$ be a group and consider $A = \text{Aut}(G)$, every $f(x) = g\cdot x$ for a fixed $g \in G$ is in $A$.
But say $G$ is a subgroup of the free group on $\Sigma$, i.e. some $H \leqslant F(\Sigma)$. Then since an automorphism in $A$ is essentially a bijection of $\Sigma$ extended homomorphically, we can count their number when $\Sigma$ finite: $|A| = |\Sigma|!$ (alphabet size factorial).
Also even "left multiplication by $g$" must always come from some permutation of the alphabet.
Example:
$$ x = aAaA $$ where $A \to aa$ (not equality!) is a what the variable would expand to in formal language theory.
Then let $g = AaA^{-1}a^{-1}$. Then $f(x) = g\cdot x = AaaA$, which is intriguing since it means grammar rule rearrangements can be done by a left multiplication action of a group. However, what permutation of the alphabet does $f(x) = g\cdot x$ correspond to?
I'm thinking I've gone wrong in my reasoning somewhere above.
First, multiplication by any fixed non-identity element of a group on either side will never be an automorphism, because it does not preserve either the identity or the multiplication.
Second, while we have an obvious injective homomorphism from $S_n$ to the automorphism group of the free group on $n$ generators, this embedding is not an isomorphism. For example, the negation automorphism of the group of integers (which is free on one generator) does not arise from a permutation of the generating set.
We also have inner automorphisms of free groups, which generally are not in the image of the embedding because any generator will commute only with its own powers in a free group.