Let $b\colon V \times V \to {\mathbb R}$ be a bilinear form. An automorphism $f\colon V \to V$ leaves $b$ invariant, iff $b(v,w) = b(f(v), f(w))$. For non-degenerate symmetric $b$ this is described by the theory of orthogonal or indefinite-orthogonal groups, depending on the signature of the bilinear form; for non-degenerate alternating / skew-symmetric $b$ there is the theory of symplectic groups.
I am interested (1) in the case of general bilinear forms which are not symmetric and not skew-symmetric (and, possibly, even not non-degenerate). (2) I am interested in a generalization for bilinear functions $\colon V\times V \to W$ into a vector space $W$.
I am probably overlooking something obvious as I did not really find any results here and it does not seem to be so complicated, after all. (I am interested in the real and complex case, so no finite fields.)
Best Answer
A bilinear form can be written as a sum of a symmetric and a skew symmetric part.
Let $c:V\times V\to \mathbb{R}: (v,w) \mapsto \tfrac12 ( b(v,w) + b(w,v) )$ and $d:V\times V \to \mathbb{R}: (v,w) \mapsto \tfrac12 ( b(v,w) - b(w,v) )$.
Then $b(v,w) = c(v,w) + d(v,w)$, so if an automorphism preserves both $c$ and $d$, then it preserves $b$. By definition of $c$, if $f$ preserves $b$, it preserves $c$ (and similarly for $d$).
So the automorphism group of $b$ is the intersection of the orthogonal automorphism group of $c$ with the symplectic automorphism group of $d$.
I found this satisfying at the time, but I believe when I tried to actually understand the intersection, I found it very difficult to compute. I was interested in finite fields, and wanted to know things like "how many elements does it have?" and I believe the answer varied a lot depending on how the groups intersected. I don't recall ever coming to a full understanding of it, but perhaps they are well known subgroups of orthogonal and symplectic groups.