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.