Does there exist an infinite finitely generated group $G$ together with a *finite* group $B$ of automorphisms of $G$ such that


1. The non-identity elements of $B$ are not inner automorphisms of $G$;
2. For every element $g \in G, g\neq 1,$ the finite set $g^B$ is a generating set for $G$?


It looks like a [Tarski Monster](http://en.wikipedia.org/wiki/Tarski_monster_group) $p$-group $T\_p$ (an infinite group in which every non-identity element has order $p$) might be a contender for $G$. If $T\_p$ has an automorphism of finite order coprime to $p$, then this would work perfectly. Unfortunately I'm not sure that it does.



Example of a finitely generated infinite group with a non-inner automorphism of finite order
I am pretty sure this example can be constructed. It might be not easy, though, since you need to have finite order automorphisms of the Tarski monsters without non-trivial fixed points. I do not remember if that has been done for Tarski monsters. For other monsters it was done, I think, by Ashot Minasyan in [Groups with finitely many conjugacy classes and their automorphisms](https://arxiv.org/abs/0704.0091). Ashot is sometimes on MO, perhaps he can give more details. If not, you can ask him directly.


I talked to Olshanskii and Osin today about this problem. They confirmed the answer. Moreover, they reminded me that there exists a paper Obraztsov, Viatcheslav N.
Embedding into groups with well-described lattices of subgroups. (English summary)
Bull. Austral. Math. Soc. 54 (1996), no. 2, 221–240 where he constructs Tarski monsters with arbitrary prescribed outer automorphism group. It is not difficult to deduce from the construction that if the prescribed outer automorphism group is, say, of order 3, then it does not have non-trivial fixed points which gives the property from the question.


Another, more straightforward, way to construct an example is the following. Fix the free group $F\_2$ and its obvious automorphism $\alpha$ of order 2 that switches the generators. Now consider the procedure of creating the Tarski monster quotient of $F$ described in the book by Olshanskii "Geometry of defining relations...". At each step together with a new relation $r\_i=1$, impose the relation $\alpha(r\_i)=1$. It is not difficult to see that $\alpha$ will be an outer automorphism of the resulting Tarski monster without non-trivial fixed points.


