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Given a group $G$, one can define the transfinite line of iterative automorphisms of $G$ to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the direct limit is taken at the limit stages:

$G\rightarrow Aut(G)\rightarrow Aut(Aut(G))\rightarrow\cdots\rightarrow G_{\alpha}\rightarrow G_{\alpha+1}\rightarrow\cdots$

The line terminates when a fixed point is reached, namely one of the groups in the chain is isomorphic to its automorphism group by the natural map. According to a result of Hamkins it is known that every automorphism line terminates. So there is no automorphism line of length $\text{Ord}$.

Definition. It is clear that the automorphism line of many non-isomorphic groups may intersect each other and so have the same terminating point. In this case, we say that two automorphism lines have converged. Otherwise, we call them parallel. Precisely, the automorphism lines of $G$ and $H$ are convergent if there are ordinals $\alpha, \beta$ such that $G_{\alpha}\cong H_{\beta}$.

Question. How many distinct parallel automorphism lines of the groups of the same cardinality do exist? Can this number vary in different forcing extensions?

Precisely, define the equivalence relation $\sim$ on (isomorphism type of) the groups so that $G\sim H$ if the automorphism lines of $G$ and $H$ converge. Let $\kappa$ be a (finite/infinite) cardinal and $\mathcal{C}_{\kappa}$ be the collection of all groups of size $\kappa$. What is the size of $\mathcal{C}_{\kappa}/\sim$ for different $\kappa$?

Remark. If an answer to the above question is in hand, the fact that every automorphism line eventually terminates actually gives us the number of groups which can arise as the terminating point of the automorphism line of a group of size $\kappa$ because two lines are parallel if and only if they have distinct terminating points.

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  • $\begingroup$ It is worth mentioning that the question seems to be closely related to finding a characterization of all fixed points of the $G\mapsto Aut(G)$ operator. Besides simple fixed points of the $Aut(-)$ operator such as the symmetric group, I am not aware of any such general characterization. $\endgroup$ – Morteza Azad Aug 14 '18 at 11:18

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