Let $\pi, \pi'$ be some Pfisterforms with $\pi \cong \pi' \perp b \pi'$ for some $b \in F^*$ and $\alpha \in D_F(\pi) = G_F(\pi) = \{ x \in F^* \mid x \pi \cong \pi \}$. So we have $\alpha \pi \cong \pi$. Is it possible to conclude $\alpha (\pi' \perp \langle b \rangle) \cong \pi' \perp \langle b \rangle$?
We assume that $F$ is a field with characteristic not equal to $2$.
A Pfisterform is a quadratic form which is isometric to $\langle 1, a_1 \rangle \otimes \cdots \otimes \langle 1, a_n\rangle$ for some $a_1, \ldots, a_n \in F^*$.
Got it. The statement is not true. Take $F = \mathbb{R}((X,Y))$, $\pi' = \langle 1, X \rangle$, $b = Y, \alpha = XY$, then whe have
$$ \alpha (\pi' \perp \langle b \rangle) \cong XY \langle 1, X, Y\rangle \cong \langle XY, Y, X\rangle \ncong \langle 1, X, Y\rangle. $$