Why is $\operatorname{SO}(p,q)$ isomorphic to $\operatorname{SO}(q,p)$? I want to prove that $\operatorname{SO}(p,q)$ isomorphic to  $\operatorname{SO}(q,p)$.
To begin with, $$\operatorname{SO}(p,q):=\{A\in M(n,\mathbb R):(Ax)^TQ(Ay)=x^TQy, \forall x,y\}=\{A\in M(n,\mathbb R):A^TQ A=Q \},$$ where the signature of $Q$ is $(p,q)$. WLOG, we may assume $Q=\operatorname{diag}(I_p, -I_q).$
There is a hint from Wikipedia I don't know how to use:

...interchanging $p$ with $q$ amounts to replacing the metric by its negative, and so gives the same group.

Any solution, using this hint or not, will be appreciated.
 A: Let $Q$ be an $n$-ary quadratic form over any field $F$of characteristic different from $2$, represented by the symmetric matrix $A$: i.e., 
$Q(x) = x^T A x$.  Then the orthogonal group of $Q$ is 
$O(Q) = \{P \in \operatorname{GL}_n(F) \mid P^T A P = A\}$
and the special orthogonal group of $Q$ is 
$\operatorname{SO}(Q) = \{P \in \operatorname{SL}_n(F) \mid P^T A P = A\}$.  
Now let $\alpha$ be a nonzero element of $F$.  Then for all $A \in \operatorname{GL}_n(F)$ (resp. in $\operatorname{SL}_n(F)$), if
$P^T A P = A$
then scaling both sides by $\alpha$ gives
$P^T (\alpha A) P = (\alpha A)$,
so $O(Q) \subset O(\alpha Q)$ (resp. $\operatorname{SO}(Q) \subset \operatorname{SO}(\alpha Q)$).  Replacing $\alpha$ by $\alpha^{-1}$ shows that we have $O(Q) = O(\alpha Q)$ and $\operatorname{SO}(Q) = \operatorname{SO}(\alpha Q)$.  
Notice that you have implicitly used that equivalent quadratic forms have conjugate (hence isomorphic) orthogonal groups.  Actually that argument is (slightly) more involved than the one above.  Combining them one sees that the conjugacy class of $O(Q)$ or $\operatorname{SO}(Q)$ depends only on the similarity class of the quadratic form.  For nondegenerate quadratic forms over $\mathbb{R}$ this amounts precisely to your statement about switching the $p$ and the $q$.    
