Pair of vectors multiplied by a random matrix and its inverse transpose are distributed randomly up to their dot product Given arbitrary nonzero vectors $\vec{x}_1, \vec{y}_1, \vec{x}_2, \vec{y}_2 \in \mathbb{Z}^{n}_p$ ($p$ prime) with $\langle x_1, y_1 \rangle = \langle x_2, y_2 \rangle$, I am trying to show that: $(R_1 \vec{x}_1, (R_1^{-1})^{T} \vec{y}_1)$ is distributed identically to $(R_2 \vec{x}_2, (R_2^{-1})^{T} \vec{y}_2)$ where $R_1, R_2$ are random invertible matrices from $\mathbb{Z}_p^{n \times n}$.
Alternatively (and actually preferred), show that for any arbitrary nonzero vectors $\vec{x}_1, \vec{y}_1, \vec{x}_2, \vec{y}_2 \in \mathbb{Z}^{n}_p$ ($p$ prime) such that $\langle x_1, y_1 \rangle = \langle x_2, y_2 \rangle$, there exists an invertible matrix $R$ such that: $R \vec{x}_1 = \vec{x}_2$ and $(R^{-1})^{T} \vec{y}_1 = \vec{y}_2$.
 A: Update: solution to the updated problem
Yes, there exists a matrix $R$.  It is easy to treat the inner product as identifying $F^n$ with its dual space and phrase the claim as follows:
Fix a field $F$ and a finite-dimensional $F$-vectorspace $V$. Let $u,v\in V$ and let $\phi,\psi\in V^*$ with all four non-zero. Then there exists $R\in\mathrm{GL}(V)$ such that $Ru = v$ and $(R^{-1})^*\phi = \psi$ iff $\phi(u) = \psi(v)$.
Proof. The condition is clearly necessary.  For sufficiency let $W = \ker(\phi)$ and let $Z = \ker(\psi)$, both spaces of codimension $1$ in $V$.


*

*Suppose $\phi(u)=\psi(v) \neq 0$. Then we have two direct sum decompositions $V = Fu\oplus W = Fv \oplus Z$. Let $R\in\mathrm{GL}(V)$ be any matrix such that $R.u = v$ and $R.W=Z$. Then $(R^{-1})^*\phi$ is a functional that vanishes on $Z$, hence is proportional to $\psi$.  But 
$\left((R^{-1})^*\phi\right)(v) = \phi(R^{-1}v)=\phi(u)=\psi(v)$ so $(R^{-1})^*\phi = \psi$.

*Suppose $\phi(u)=\psi(v) = 0$. Then $u\in Z,\,v\in W$. Let $z,w \in V$ be vectors such that $\phi(z)=\psi(w)=1$.  Now choose $R\in\mathrm{GL}(V)$ such that $R.Z=W$ and such that $Ru = v$ and  $Rz = w$. The second choice is possible since an isomorphism of $Z,W$ can be chosen to map any specific non-zero vector to any specific non-zero vector, the second since $z,w$ are linearly independent  of $Z,W$ respectively. By construction we have $Ru=v$ and $R.Z = W$, the latter forcing $(R^{-1})^*\phi$ to be proportional to $\psi$.  Finally, $Rz = w$ ensures that $(R^{-1})^*\phi = \psi$ and we are done.
Solution to the original problem
The original statement concerned the distribution of the pair $\left(Rx,R^{-1}y\right)$ rather than $\left(Rx,(R^{-1})^Ty\right)$.  The distribution is as follows:
Let $R$ be a uniform invertible matrix. The marginal on $\vec{z} = R \vec{x}$ is then a uniform non-zero vector.  Conditioning on $\vec{z}$ we have two possibilities: either $\vec{z}$ is proportional to $\vec{y}$, at which point $R^{-1}\vec{y}$ is a fixed multiple of $\vec{x}$, or it is not, at which point $R^{-1}\vec{y}$ is uniformly distributed over all vectors not proportional to $\vec{x}$.
It follows that the pair $\left(R\vec{x},R^{-1}\vec{y}\right)$ is distributed over the union $$ \left\{ \left(\vec{z},\vec{w}\right) \mid \vec{z}\textrm{ indep of }\vec{y} \wedge \vec{w}\textrm{ indep of }\vec{x}\right\} \cup \left\{ \left(\alpha\vec{y},\alpha^{-1}\vec{x}\right) \mid \alpha\in\mathbb{F}_p^\times\right\} $$
where the distribution is uniform in each side, and the probability of ending on the right is $\frac{p-1}{p^n-1} = \frac{1}{1+p+\cdots+p^{n-1}}$.
Conclusion: in the original formulation


*

*The distribution depends on the pair $\left(\vec{x},\vec{y}\right)$.

*The pairing $\left\langle<\vec{x},\vec{y} \right\rangle$ does not play a role.

