So this question is a continuation of the following one
For some motivations and relevant backgrounds related to this question see K. Conrad's answer in [1].
So let $K$ be a field of characteristic $2$ and let $V$ be an $n$-dimensional vector space over $K$. Recall that a function $Q:V\rightarrow K$ is a quadratic form on $V$ if
(1) $Q(av)=a^2v$ for all $a\in K$ and $v\in V$,
(2) $(v,w)\mapsto B(v,w):=(Q(v+w)-Q(v)-Q(w))$ is a $K$-bilinear form.
Define $O(Q):=\{A\in GL(V):Q(Av)=Q(v)\;\;\mbox{for all $v\in V$}\}$ to be the isotropy group of $Q$ (or isometry group of $Q$).
We say that $Q$ is non-degenerate if $Q(v)=0$ and $B(v,V)=0$ imply that $v=0$.
Q: If $Q$ and $Q'$ are non-degenerate quadratic forms on $V$ and $O(Q)=O(Q')$ does this imply that there exists $\lambda\in K^{\times}$ such that $Q=\lambda Q'$ (in such a case we say that $Q$ and $Q'$ are associated)?
Note that the answer is positive if $char(K)\neq 2$. So you may try to complete K. Conrad's argument (which breaks down when $char(K)=2$) or you may try to provide an example of two non-degenerate non-associated quadratic forms $Q$ and $Q'$ such that $O(Q)=O(Q')$.
