Is there a finite non-abelian group $G$ of prime exponent such that the full automorphism group of $G$ is of odd order? 



The parity of the full automorphism group order of finite non-abelian groups of prime exponent
Yes:



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> for every prime $p\ge 7$, there's a finite group of exponent $p$ whose automorphism group is a $p$-group.
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**Initial answer** *(Jan 18' 2016)*
Start from any (finite-dimensional) complex nilpotent Lie algebra $\mathfrak{g}$ that is defined over $\mathbf{Q}$ and has a unipotent automorphism group (see e.g. Luks' Lie algebra top of p14 in [(Ancochea Campoamor survey)](http://arxiv.org/abs/math/0011224) for a 16-dimensional example), see also first edit (Jan 19' 16) below.


Fix a rational structure, so we can defined $\mathfrak{g}\_p$ as the corresponding Lie algebra modulo $p$ for $p$ large enough (i.e., not dividing any of the denominators). (Formally speaking, this means writing $\mathbf{g}=\mathfrak{h}\otimes\_{\mathbf{Q}}\mathbf{C}$ for some Lie algebra $\mathfrak{h}$ over $\mathbf{Q}$.)


Then for $p$ large enough, the automorphism group of $\mathfrak{g}\_p$ is a $p$-group. Indeed otherwise, if for a growing subsequence $(p\_i)$ this is not the case, say $\mathfrak{g}\_{p\_i}$ has an automorphism of prime order $\neq p\_i$, passing to the algebraic closure $K\_i$ of $\mathbf{F}\_{p\_i}$ and then to a ultralimit over $i$ we get that the ultraproduct, which is a Lie algebra over the ultraproduct $K$ of the $K\_i$ has a nontrivial semisimple automorphism. But this ultraproduct is isomorphic to $\mathfrak{h}\otimes\_{\mathbf{Q}}K$. This contradicts that $\mathfrak{g}$ has a unipotent automorphism group.


Note that this provides a group of order $p^{16}$ with some well-controlled nilpotency length (I haven't made the computation for Luks' Lie algebra but it's easy to do), but gives no control of $p$, although for small $p$ a computer computation is doable).




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We can expect a computer computation to yield explicit examples.


Anyway no example is possible for $p$-group of nilpotency length 2 (and odd exponent $p$): indeed such a group admits an action by automorphisms of the multiplicative group modulo $p$ (of order $p-1$), because the corresponding Lie algebra is Carnot. Classification along with Malcev correspondence also shows that no group of order $p^{\le 6}$ for prime $p\ge 7$ can work, because the corresponding Lie algebras have nontrivial integral gradings according to the classification by de Graaf in [here](http://www.sciencedirect.com/science/article/pii/S0021869306005254) (J. Algebra 2007) [this also works in order $5^{\le 5}$, and in order $5^6$ except for nilpotency length $=5$].




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**Edit:** *(Jan 19' 2016)* I have checked that the 7-dimensional (with nilpotency length 6) Lie algebra given page 17 line 8 in [(Ancochea Campoamor survey)](http://arxiv.org/abs/math/0011224), viewed with coefficients in an arbitrary field, has only unipotent automorphisms. If one takes it in the field on $p$ elements for any prime $p\ge 7$, the Baker-Campbell-Hausdorff (which realizes Malcev's correspondence) provides a group of order $p^7$ whose automorphism group is a $p$-group (thus of odd order).


According to what's above and computer check by Alireza for $p=3,5$, for $p$ odd any nontrivial $p$-group of order $\le p^6$ has an automorphism of order 2. So $p^7$ is the smallest possible order for such exceptions, for $p=3,5$ possibly one has to go a little higher since Alireza checks that every group of order $3^7$ has an automorphism of order 2. (At the moment this gives no example for $p=3,5$ although they certainly exist.)




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**Edit:** *(Jan 25' 2018)* Thinking a little about the remaining cases $p=3,5$, they seem different and deserve separate additional comments:


first recall by above remark that for any odd $p$, the nilpotency length of any example should be $\ge 3$.


* for $p=5$, Malcev correspondence is valid in nilpotency length $\le 4$. Therefore if one searches among nilpotent rational Lie algebras of nilpotency length in $\{3,4\}$ and unipotent automorphism groups, one can reasonably hope to find one that reduces modulo 5 and for which the reduction modulo 5 has no extra automorphisms.
* $p=3$ is quite different, because exponent 3 implies nilpotency length $\le 3$ (in contrast, by Razmyslov's theorem, for all $p>3$ there are finite groups of exponent $p$ and arbitrary large solvability length). So in this case, Lie algebras are of no help. Finding a finite group of exponent 3 whose automorphism group is a 3-group (or of odd order) sounds a quite different question.


