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The same post's presented on mathstackexchange, but I think this one is a research level question.

I was preparing for my postgraduate exams and one of the questions concerned constructing non-isomorphic boolean algebras with isomorphic automorphism groups. I know that it's proved using Stone's representation and a fact from topology (that there are actually spaces, satisfying the requirements). It's natural to expect those algebras to exist, because it's too strong condition to be fully described by own automorphism group.

So after learning it the question arouse: why would anyone construct such algebras? More generally, are there any (algebraic) structures that are solely defined by their own automorphism group? Formally let $A,B$ be algebraic structures and $\text{Aut}(A), \text{Aut}(B)$ be their automorphism groups. For which classes of structures it follows: $$\tag{1}\label{1}\text{Aut}(A) \cong \text{Aut}(B) \implies A \cong B \; ?$$

It's relatively obvious, that non-empty finite sets satisfy the requirements (as noted on mse). It's a bit tricky, but we can show that it's still true for any non-empty sets. Are there any other non-trivial examples of this situation. Can we deduce any properties of those structures, satisfying $\eqref{1}$? Probably Aut group must be big comparing to the structure itself (if we use the analogy from Sets).

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    $\begingroup$ Amritanshu Prasad showed that this also holds for vector spaces in an answer to one of my early questions. $\endgroup$ Commented Aug 27 at 15:26
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    $\begingroup$ @LSpice Doesn’t the answer there only show one can obtain the dimension of $V$ from certain properties of the action of $GL(V)$ on $V$, not just the group $GL(V)$? I’m not sure how to state “there exists $v_1, \cdots, v_n \in V$ s.t. …” internally within $GL(V)$. $\endgroup$ Commented Aug 27 at 15:55
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    $\begingroup$ @DavidGao, re, yes, there are a multitude of ways that one could ask the question. In the context in which I asked, I was considering identifying the actual subgroup of the permutation group consisting of linear automorphisms, not just the isomorphism type of $\operatorname{GL}(V)$. But you are right that Vladimir Dotsenko's and Anton Klyachko's answers are more in the spirit of this question. $\endgroup$ Commented Aug 27 at 16:11
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    $\begingroup$ In the absence of choice it's possible for there to be infinite-dimensional vector spaces $V$ over a field $K$ such that $\text{End}(V) \cong K$, hence such that $\text{Aut}(V) \cong K^{\times}$, see math.stackexchange.com/questions/5087218/…. $\endgroup$ Commented Aug 27 at 16:15
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    $\begingroup$ @LSpice Thanks for the clarification! $\endgroup$ Commented Aug 27 at 16:16

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Abelian $p$-groups are determined by their automorphism groups.

One may object that this category is not a category of algebraic structures bona fide; but nevertheless, it is a locally finitely presentable (Grothendieck abelian) category, which kinda qualifies, in my opinion.

The proof is somewhat intricate; there's no full published journal version.

Technical core of this theorem is the fact that $p$-group $A$ is determined by (non-unital) subring $\Delta \subset \mathrm{End}(A)$, consisting of finite order elements of Jacobson radical; it's written in [1].

Lie-theoretical methods allow to determine $1 + \Delta \subset \mathrm{Aut}(A)$ internally (without reference to $A$ or endomorphism ring); this is written in proceedings of conference commemorating A. Orsatti's 60th birthday [2].

[1] J. Hausen, C. E. Praeger, and P. Schultz, Most Abelian p-groups are determined by the Jacobson radical of their endomorphism rings, Math. Z. 216 (1994), 431-436.

[2] P. Schultz, Automorphisms which determine an abelian p-group, Abelian groups, module theory and topology, 1998, pp. 373-379, https://doi.org/10.1201/9780429187605

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