Odd-order p-groups with as many automorphisms as elements

Is there a finite group of odd prime-power order whose automorphism group has exactly as many elements as the group itself? Abelian examples are impossible, and for decades it was even conjectured that the group's order always divides the number of automorphisms. In 2026 a computer-assisted construction proposed an example of order $1009^{52359}$ , which awaits independent verification.

Problem

Let $p$ be an odd prime. Does there exist a non-trivial finite $p$ -group $G$ with

|\mathrm{Aut}(G)| = |G|?

Since $|G|$ is then a power of $p$ , such a $G$ must in particular have no automorphism of order prime to $p$ . For example, no non-trivial abelian group of odd order works: inversion $x \mapsto x^{-1}$ is an automorphism of order 2.

Background

The question was posed by Des MacHale in the 16th issue of the Notebook (2006). It sits next to a well-known older problem (Archive 12.77, A. I. Starostin): does the order of a finite non-cyclic $p$ -group of order greater than $p^2$ always divide the order of its automorphism group? González-Sánchez and Jaikin-Zapirain (2015) answered that negatively, constructing for each prime $p$ families of $p$ -groups with $|\mathrm{Aut}(G)|/|G| \to 0$ . Equality $|\mathrm{Aut}(G)| = |G|$ is a different, delicate balance between these regimes.

What is known

Status in the Notebook (marked ♯). The Notebook marks this problem with ♯: an artificial-intelligence system has proposed a solution that mathematicians have not yet confirmed, so the problem is to be regarded as unsolved. Readers are invited to confirm the solution or find mistakes.
The proposed solution. The cited preprint (A. Kriebel, version of 15 September 2026) answers yes with $p = 1009$ and $|G| = |\mathrm{Aut}(G)| = 1009^{52359}$ . The group is obtained through the Lazard correspondence from an explicit rank-31 Lie ring (based on a Lie algebra with only inner derivations, due to Omirov and Ruan) taken modulo $1009^{1689}$ ; its nilpotency class is at most 846, below $p$ . Counting automorphisms is reduced to the kernel of an explicit $14415 \times 961$ integer matrix, whose Smith normal form invariants were computed with independent Python and C++ programs. The author states that the construction and proof were generated with AI tools, and that the checks are neither external peer review nor a proof-assistant formalisation.
Partial results. Shabani-Attar (2015) excluded odd-prime examples among $p$ -groups of maximal class, groups with cyclic Frattini subgroup, and groups of class two with cyclic centre. For every prime $p$ , no group of order $p^6$ has an automorphism group of order $p^6$ (Yu, Ban and Zhang, 1996; Notebook Archive 12.78).

What would count as progress

Verify or refute the proposed solution: check the theoretical steps (the finite-field stabiliser computation, the lifting and logarithm–exponential arguments, and the Lazard correspondence), and rerun the published certificates for the Smith invariants with independent software.
A smaller or more transparent example, ideally of a size where $|\mathrm{Aut}(G)|$ can be computed directly in GAP or Magma. Exhaustive searches through small orders for $p = 3, 5, 7$ can rule out small cases.
A formal verification of the counting argument for the proposed group.

References

J. González-Sánchez and A. Jaikin-Zapirain, Finite p-groups with small automorphism group, Forum Math. Sigma 3 (2015), e7; arXiv:1406.5772.
M. Shabani-Attar, On equality of order of a finite p-group and order of its automorphism group, Bull. Malays. Math. Sci. Soc. 38 (2015), 461–466.
A. Kriebel, An explicit odd-order p-group with as many automorphisms as elements, preprint, version 1.1.0 (2026), as cited in the Notebook.

Source

Imported from The Kourovka Notebook 16.63, curated by E. I. Khukhro and V. D. Mazurov. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.