Let $p$ be a prime, $FD_p$ the finite field of order $p$, $n\ge 3$, $V=(F_p)^n$ and let $f:V \to V$ be a bijection such that the dot products $u \cdot v$ and $f(u) \cdot f(v)$ are either both zero or both non-zero for all $u,v \in V$. Is $f$ a linear function on $V$?
I am just using the standard dot product on $V$ relative to some fixed basis.
The result is true for small values of $n$ and $p$, as can be checked using, for example, Magma.
I am expecting that this is a standard result in finite projective geometry, but have yet to find a reference.
Let $PV=\{a_1,\cdots,a_k\}$ be the corresponding projective space where $k=|PV|$. Let $M$ be the $k \times k$ matrix over $\mathbb Q$ with the $i,j$ entry $0$ if and only if $a_i \cdot a_j =0$, otherwise it is $1\in \mathbb Q$.
Then $S_k \times S_k$ acts by permuting the rows and columns of $M$. Also $G=SL(n,p)$ acts on $V$ and $V^*$ giving a permutation action on the rows and columns of $M$ (respectively). Thus there is a homomorphism $G \to S_k \times S_k$ that is injective. The dot product is just the action of $V^*$ on $V$ in this setup.
So the problem is equivalent to showing that the centralizer in $S_k \times S_k$ of $M$ is the image of $G$ in $S_k \times S_k$.
It's not true, and here's a counterexample. Let $V = F_3^3$. Define $f(x, y, z)$ by $$f(x, y, z) = \begin{cases} (-x, y, z) & \text{if } y = z = 0 \\(x, y, z) & \text{otherwise.} \end{cases}$$ This map stabilises (permutes, even) one-dimensional subspaces (lines through the origin). Orthogonality is a property not just of pairs of vectors, but pairs of one-dimensional subspaces, as the dot product is bilinear. As such, any pair of orthogonal one-dimensional subspaces will map to the same pair of orthogonal one-dimensional subspaces, and so orthogonality is preserved.
However, the function is certainly not linear. For example, $$f(1, 0, 0) + f(0, 1, 0) = (2, 0, 0) + (0, 1, 0) \neq (1, 1, 0) = f(1, 1, 0).$$