When does there exist an isometry that switches two subspaces? Let $V$ be a real vector space of finite dimension and let 
$\langle \cdot, \cdot \rangle$ be a non-degenerate symmetric 
bilinear form on $V$. Let $U, W \subseteq V$ be linear 
subspaces such that the bilinear forms $\langle \cdot, \cdot 
\rangle \vert_U$ and $\langle \cdot, \cdot \rangle \vert_W$ 
are isometric and non-degenerate.
It is quite easy to prove that there exists an isometry $f$ 
of $V$ such that $f(U) = W$. I would like to know whether 
there are some sufficient conditions such that there exists 
an isometry $f$ of $V$ such that $f(U) = W$ and $f(W) = U$.
Thanks to all! 
EDIT. Since I have not received any answer, I want to 
say that I'm interested in a very particular case: $V = 
\mathbb{R}^{n+1}$, $\langle \cdot , \cdot \rangle$ is the 
Lorentzian scalar product on $\mathbb{R}^{n+1}$, i.e. 
$\langle x, y \rangle = \sum_{i=1}^n x_i y_i - x_{n+1} 
y_{n+1}$, and $U$ and $W$ are $2$-dimensional subspaces of 
$\mathbb R^{n+1}$ such that $\langle \cdot, \cdot
\rangle \vert_U$ and $\langle \cdot, \cdot \rangle \vert_W$ 
have signature $(1,1)$. 
 A: If the scalar product


*

*restricted to $U$ and $V$ is definite and has the same sign, and

*is not degenerate on $U+V$


then the answer is yes.  This also applies to some of your cases since you can switch the orthogonal complements $U^{\perp}$ and $V^{\perp}$.  It would be nice if the second condition could be dropped, but I'll need to think about that some more.
Let $\pi_U$ and $\pi_V$ be the orthogonal projections onto $U$ and $V$ with respect to the scalar product.  Then $\pi_U \pi_V$ maps $U$ into $U$ and its restriction to $U$ is self adjoint.  That is, for all $v, w \in U$
$$
\langle \pi_U\pi_V v, w \rangle = \langle v, \pi_U\pi_V w \rangle.
$$
This means that $U$ has an orthonormal basis $u_1, \dotsc, u_k$ of eigenvectors for eigenvalues $0 \leq \lambda_1, \dotsc, \lambda_k \leq 1$.  Likewise, $V$ has an orthonormal basis of eigenvectors $v_1, \dotsc, v_k$ for the same eigenvalues.  That the eigenvalues are the same follows from
$$
\pi_V\pi_U(\pi_V u_j) = \pi_V(\pi_U\pi_V u_j) = \lambda_j \pi_Vu_j
$$
and
$$
\langle \pi_Vu_j, \pi_Vu_j \rangle = \langle \pi_Vu_j, u_j \rangle = \langle \pi_Vu_j, \pi_Uu_j \rangle = \langle \pi_U\pi_Vu_j, u_j \rangle = \lambda_j.
$$
In fact, for any eigenvalue $\lambda_j \neq 0$ the projection $\pi_V$ is $\lambda_j^{1/2}$ times an isometry from the eigenspace in $U$ onto the eigenspace in $V$.  In particular we can chose the basis $(v_j)$ such that $\pi_Vu_j = \lambda_j^{1/2}v_j$ for all $j$.
Now define $f$ by
$$
f(x) = \begin{cases}
v_j & \textrm{if }x = u_j\\
u_j & \textrm{if }x = v_j\\
x & \textrm{if } x \in U^{\perp} \cap V^{\perp}
\end{cases}.
$$
Then $f$ switches the subspaces $U$ and $V$ and by the observations above one can check that it is also an isometry.
