Let $G$ be a finite $p$- group and let $\varphi$ be an automorphism of $\mathbb{F}\_pG$ as $\mathbb{F}\_p$-algebras and let $n = \sum\_{g\in G} g$ be the norm element. Does it follow that $\varphi(n)=n$?


Since the $\mathbb{F}\_p$-vector space spanned by $n$ is exactly the annihilator of the maximal ideal, it follows that this vector space is characteristic and thus $\varphi(n)=\lambda n$ for some $\lambda \in \mathbb{F}\_p^\*$ and thus we have a group homomorphism:
$Aut(\mathbb{F}\_pG)\rightarrow \mathbb{F}\_p^\*$. So the question is whether this map is always trivial.



Is the norm element characteristic in modular group rings?
I know this is an old question, but I think the answer is yes. The idea is to pass to the associated graded for the Jennings filtration, and look at the $p$-restricted Lie algebra $L(G)$ you get that way. The element $n$ corresponds to the socle element in the restricted universal enveloping algebra, which is one dimensional. Taking a basis of homogeneous elements, $x\_1,\dotsc,x\_m$, of $L(G)$, the norm element is $(x\_1\dotsm x\_m)^{p-1}$. An automorphism of $G$ acts as a linear transformation on $L(G)$, and multiplies $x\_1\dotsm x\_m$ by a constant that depends on that linear transformation (modulo commutators further down the filtration, and squares), and this constant is an element of $\mathbb{F}\_p^\*$. Taking $(p-1)$st powers, the action is trivial on $n$, because those commutators and squares drop off the end. This argument needs checking, but I think it's right.


Edit 20 March 2023: I found a clean way to present the proof, which can be found at [The socle of the group algebra of a finite $p$-group](https://arxiv.org/abs/2303.09940).


