For a fixed positive diagonal matrix $D$, the set of all real matrices $A$ satisfying $A^T = -D A D^{-1}$ form a Lie algebra with the matrix commutator as the Lie bracket.
Since
$$
\left( e^{A} \right)^T = e^{A^T} = e^{-D A D^{-1}} = D e^{-A} D^{-1},
$$
the Lie group consists of matrices $G=e^{A}$ satisfying
$$
G^T = D G^{-1} D^{-1} \implies G = (D^{-1})^T (G^{-1})^T D^T
$$
Does this Lie group have a special name?
Remarks:
1) The Lie algebra corresponds to ODE's whose trajectories are ellipsoids. Please see the post When do the solutions of a linear ODE system lie on ellipses?.
2) Note that in the special case of $D$ being the identity matrix, this Lie algebra is the skew-symmetric matrices and the corresponding Lie group is $SO(n)$, the group of rotations.
Best Answer
In the following, $^t A$ means the transpose of $A$, in particular for a column vector $x= \pmatrix{x_1\\\vdots\\x_n}$ we have $^t x = (x_1 \dots x_n)$.
Any symmetric matrix $D \in M_n(\mathbb R)$ defines a symmetric bilinear form $b_D$ on $\mathbb R^n$ via $b_D(x,y) := (^tx) Dy$. Check that your relation says that exponentials of elements of your Lie algebra are elements $G \in GL_n(\mathbb R)$ which are invariant with respect to $b_D$, i.e. $b_D(Gx, Gy) = b_D(x,y)$. Elements of this kind form the orthogonal group with respect to $b_D$ (or $D$, or the corresponding quadratic from $q_D(x) := b_D(x,x)$). Since a Lie group is only locally determined by its Lie algebra, you actually might have various Lie groups sitting above that Lie algebra. Call the Lie algebra $\mathfrak{so}_{n, D}$ for the moment.
In the case $D = Id$, you've recovered the standard orthogonal groups $O(n)$, or $SO(n)$ (or their covers $Pin(n)$ or $Spin(n)$) with the Lie algebra $\mathfrak{so}_n$.
Now everyone has learned Gram-Schmidt orthonormalisation and Sylvester's Law of Inertia which says that over $\mathbb R$, such a quadratic form is actually up to equivalence (i.e. base change) determined by its signature. Further, if the base change is given by a matrix $P$, i.e. $^t P D_1 P = D_2$, then check that $A \mapsto P^{-1}A P$ (NB: now one really takes the inverse, not the transpose) defines an isomorphism $\mathfrak{so}_{n, D_1} \simeq \mathfrak{so}_{n, D_2}$.
In particular, since in your setting $D$ is positive definite, there is a base change matrix $P$ (indeed, for $D = diag(a_1, ..., a_n)$ just scaling the $i$-th coordinate with $\sqrt a_i$) such that $^t P D_1 P = Id$, and your Lie algebra is indeed isomorphic to the standard special orthogonal Lie algebra, and the Lie groups sitting above it are isomorphic to the standard ones mentioned above.
Note that if instead of ellipses and corresponding quadratic forms with all $a_i$ positive, you had hyperbolas and some of the $a_i$ negative (correspondingly, "hyperbolic" i.e. isotropic subspaces in your quadratic form $q_D$), then instead you would get Lie algebras of one of the indefinite orthogonal groups.