Let $K$ be a field extension of $F$. If two $n\times n$ matrices $A,B \in M_n(F)$ are similar via a matrix $P \in GL_n(K)$ (that is, $A=PBP^{1}$), then there exists a matrix $Q\in GL_n(F)$ such that $A=QBQ^{1}$. The question is: if $P$ is symmetric, can $Q$ be chosen to be symmetric?
Yes, if $F$ is infinite. The set of symmetric matrices $Q$ such that $AQ=QB$ is a vector space $V$ over $F$. We know that $V\otimes _FK\neq 0$, hence $V\neq 0$. The polynomial $\det_{F}$ is not identically zero in $V\otimes _FK$, hence in $V$ since $F$ is infinite. Thus there exists $Q\in V$ invertible.


1

1$\begingroup$ If $F$ is finite and we have to find a point $a$ in which, for some polynomial $f$, $f(a)\ne 0$, it may appear that it exists in $K$, but not in $F$. Thus even if it is true, it has to be studied carefully. $\endgroup$ – Fedor Petrov Feb 3 '16 at 9:03

1

1$\begingroup$ No — actually I don't have much feeling whether a counterexample exists or not. $\endgroup$ – abx Feb 3 '16 at 11:06