How does changing a bilinear form change its associated orthogonal group? I'm trying to understand bilinear forms and how they can be 'equivalent' and how this 'equivalence' translates to their associated orthogonal groups.
Fix a field $F$. I'm comfortable translating between a quadratic form $Q$ (homogenous degree $2$ polynomial in $\ell$ variables), its bilinear form on $F^\ell$ $B_Q$, and the symmetric $\ell \times \ell$ matrix $A_Q$. Each of these 'representations' has a group of matrices which preserve them, for example $SO(Q,F) := \{g \in SL_(\ell,F)  |  Q(gv) = Q(v) \; \forall v \in F^\ell\}$. Each of these representations has an associated group of matrices preserving it and I can show these groups are literally equal.
On the other hand, there are operations on the vector space that seem to change the bilinear form into things more pleasant to deal with ('diagonalizing the quadratic form') and some notation ($SO(p,q)$) that seems to indicate that all of these are determined by the 'signature' of the quadratic form, etc. It seems like this should be explained by changing the underlying basis for the vector space, but I can't quite see it.
Since my context is reading about these orthogonal groups as Lie groups (Morris Arithmetic Groups), I'd like to know how changing the form affects the group and why. It seems like it should be conjugation by a change of basis matrix, but again, I can't really explain why or find a reference.
Explicitly, I'm looking for an explanation of how/when quadratic forms can be considered 'equivalent' (for example, why does the signature characterize them?), what effect that has on the orthogonal groups, and hopefully how it translates between the various representations listed above. I know that's a lot, so a reference would also be good.
 A: You say you can show all these groups are the same, which certainly is not true set theoretically, but in fact is not even true group theoretically when one interprets "same" as "isomorphic".
Consider these two examples:

*

*$F=\mathbb{R}$, $l=2$, with $A_Q = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

*$F=\mathbb{R}$, $l=2$, $A_Q = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.

In case 1, as you noted one gets the (special) orthogonal group $SO(2;\mathbb R)$. This is isomorphic to the circle group $S^1 = \{z \in \mathbb C^{\times} \bigm| |z|=1\}$. One sees from this, for example, that the orthogonal group has elements of all finite orders $n=2,3,4,\ldots$, namely the matrices
$$\begin{pmatrix} \cos(2\pi/n) & \sin(2\pi/n) \\ -\sin(2\pi/n) & \cos(2\pi/n)  \end{pmatrix}
$$
The geometric point to ponder is that the level sets of the norm are circles centered on the origin, and the group action must preserve each circle (and must rotate that circle rigidly)
But in case 2 one gets instead the group $SO(1,1;\mathbb R)$. That group does have a few elements of bounded finite order, but it has an index $4$ normal subgroup isomorphic to $\mathbb R$; you can do a basis change so that this subgroup becomes
$$\begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix}
$$
It follows that the possible finite orders of group elements are bounded by $4$. In this case the geometric point to ponder is that level sets of the norm are (2-sheeted) hyperbolas, and the group action must preserve the 4 hyperbolas formed as the union of the norm $1$ and norm $-1$ level sets.
So the two groups cannot be isomorphic as abstract groups. It's even worse than that: When considered as topological groups or as Lie groups, $SO(2;\mathbb R)$ is compact whereas $SO(1,1;\mathbb R)$ is noncompact.
It's definitely worth pondering how these groups can be classified up to isomorphism as Lie groups, or as topological groups (or even just as abstract groups). Here the absolute value of the signature  plays an important role as an invariant.
