Let $A$ be a finite dimensional , associative, unital $F$-algebra, where $F$ is a field.


Let $s\_A:A\to F$ be an $F$-linear map. 
Now consider an arbitrary field extension $K/F$, and define $s\_{A\otimes\_FK}$ as the unique $K$-linear map such that $s\_{A\otimes\_F K}(a\otimes 1)=s\_A(a) $ for all $a\in A$.


**Question.** Assume that for all automorphism $f:A\to A$ of $F$-algebras, we have $s\_A\circ f=s\_A$. 


Do we have the same property for $A\otimes\_FK$, that is that for all automorphism $g:A\otimes\_FK\to A\otimes\_FK$ of $K$-algebras, $s\_{A\otimes\_F K}\circ g=s\_{A\otimes\_FK}$ ?


If the answer is no , can we find necessary and sufficient conditions on $s\_A$ for this to be true ? 


The answer is positive for all examples I know (separable algebras with the reduced traces, plus few more non separable examples), but I have no clue whether it is true or not , and how to proceed to prove it if it is indeed true.


I will be happy to assume that $F$ is infinite if this is necessary.


I am also interested in the same question when $s\_A$ is replaced by a bilinear form over $A$, but I suspect that the answer would be the same.



Linear maps preserved by algebra automorphisms
Here's an example where the answer is no ($d+1$-dimensional for $d\ge 2$). Let $L$ be a separable extension of $F$ of finite degree $d\ge 2$ (we assume this exist, i.e. $F$ is not separably closed), and assume that $L$ splits over $K$ (e.g., assume that $K$ is separably closed). Say, $F=\mathbf{R}$ and $L=K=\mathbf{C}$ if this language is not familiar.


Consider $A=F\times L$ as $A$-algebra, and $s\_F$ is the projection from $A$ to $F$ with kernel $L$ ($s\_F$ is a more accurate notation than $s\_A$ since I can write $s\_K=s\_F\otimes\_F K$). Since $\mathrm{Aut}(A)\_F$ consists of the $$F\times L\ni (t,u)\mapsto (t,\alpha(u)), \quad \alpha\in\mathrm{Aut}\_F(L),$$
we indeed have $s\_F\circ f=s\_F$ for every $f\in\mathrm{Aut}(A)\_F$.


But on the "complexification" $A\otimes\_FK$ is isomorphic to $K^{d+1}$ as $K$-algebra, and in particular all the symmetric group $\mathfrak{S}\_{d+1}$ acts by permuting, and the set of linear forms commuting with all automorphism reduced to the "sum of coordinates" map, which vanishes on no nonzero idempotent. Since $s\_F$ vanished on the nonzero idempotent $1\_L$, we deduce that $s\_K$ does not commute with all automorphisms.


