Let $h\colon \mathbb{C}^{2n} \times \mathbb{C}^{2n} \rightarrow \mathbb{C}$ be the form $$h(\vec{v},\vec{w}) = \left(v_1 \overline{w}_1 + \cdots + v_n \overline{w}_n\right) - \left(v_{n+1} \overline{w}_{n+1} + \cdots + v_{2n} \overline{w}_{2n}\right).$$ Define $G$ to be the subgroup of $\text{GL}(2n,\mathbb{C})$ stabilizing $h$, i.e. $G$ is all invertible $2n \times 2n$ complex matrices $M$ such that $h(M \vec{v}, M \vec{w}) = h(\vec{v},\vec{w})$ for all $\vec{v},\vec{w} \in \mathbb{C}^{2n}$. Equivalently, $M$ should satisfy $$M^t \left(\begin{matrix} \mathbb{1}_n & 0 \\ 0 & -\mathbb{1}_n \end{matrix}\right) \overline{M} = \left(\begin{matrix} \mathbb{1}_n & 0 \\ 0 & -\mathbb{1}_n \end{matrix}\right).$$ Question: which real Lie group is $G$? I assume it is reductive, but I have trouble identifying it with a product of any of the standard reductive real Lie groups. But my research is usually in infinite discrete groups, so I might just be ignorant of standard Lie group stuff.
I'd also be interested in the more general question where you distribute the $\pm 1$'s along the diagonal of the matrix defining $h$ differently (and thus don't insist that the dimension be even). If the matrix is just the identity, then you get the usual unitary group.