Let $V \in R^{d \times d} $ be a real orthogonal matrix. Denote by $V \circ V$ the Hadamard product (elementwise product).
I wish to show that if $V \left(V \circ V \right)^T V$ is "close" to $V \circ V$, then $V$ is necessarily "close" to $V \circ V \circ V$, meaning $V$ is an orthogonal matrix with entries in $\{-1, 0, 1\}$ (a permutation matrix upto the sign of the entries).
In precise terms, I want to prove that:
$V \left(V \circ V \right)^T V = V \circ V \implies V_{ij} \in \{-1,0,1\}$. So, equality implies $V$ is a permutation matrix up to the sign of the entries.
For a given $\epsilon > 0$ to prove there exists an $\hat{\epsilon} >0$ such that: $||V \left(V \circ V \right)^T V - V \circ V || < \hat{\epsilon} \implies ||V- V\circ V \circ V|| < \epsilon $. Meaning that approximate equality implies $V$ is close to a permutation matrix (upto the sign). The norm is any standard matrix norm.
I don't think that's true. Let $$ V=\begin{bmatrix}1&0\\0&-1\end{bmatrix}. $$ Then $V\circ V=I_2$, and $$ V(V\circ V)^TV=V^2=I_2=V\circ V. $$