Let $s \in \Sigma^\*$ be a formal language string. Consider the automorphism group of $s$, defined to be the set of all permutations of positions of $s$ that leave $s$ fixed. For instance $G(abab) = \{\text{id}, \ (1,3), \ (2,4), \ (1,3)(2,4)\}$. 


What about automorphisms of the associated grammars of strings. For instance:


$$ g = \\
A \to BB \\
B \to ab
$$


has automorphism group $\langle (1,2) \rangle$ (you simply re-index the symbols however you want and compute the group that fixes a *set* of strings which are the rule RHS's of the grammar).


Have you heard of this conjecture or a reformulation of it?:


*The automorphism groups of the smallest grammars of a string are isomorphic to one another.*


An example
----------


$$
s = abc \ abc \ abc \ ab
$$. 


Positions are $1, 2, \dots, |s|$. The symmetry group of the string acts on this position set as a subset of $S\_{|s|}$. Find the symmetry group as $G(s) = \langle \text{id}, (1,4), (2,5), (3,6), (1,7),(2,8),(3,9),(1,10), (2,11) \rangle$. Now let's choose a grammar of $s$ that's strictly smaller than $|s|$. How about 


$$g = \\
A \to BB abc ab \\
B \to abc
$$


Then order the positions accordingly to the order: BB abc ab, abc. Then the symmetry group is $G(g) = \langle (1,2), (3,6), (3,8),(4, 7),(4,9),(5,10)\rangle$. 



The automorphism groups of smallest grammars of a language string are isomorphic
I think the grammars $A \to aaaaa$ and $A \to BaB$, $B \to aa$ are both minimal (5 symbols). 


If I understand your definitions correctly, the automorphism group of the first one is $S\_5$ with $120$ elements, the automorphism group of the second is $S\_2\times S\_3$ with $12$ elements.


