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Yes. 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 invertible matrices form a Zariski open subset $U$ of $V$ (defined by $\det\neq 0$). Again we know that $U\otimes _FK$ is a nonempty Zariski open subset of $V\otimes _FK$; since $V$ is Zariski dense in $V\otimes _FK$, $U$ is nonempty.