$E\subset\mathbb{R}^n$ is an ellipsoid if $E = E(g):= \{x\in \mathbb{R}^n \mid x^t g x \le 1\}$ for some inner product $g$ on $\mathbb{R}^n$. Given an ellipsoid $E\subset\mathbb{R}^n$, how unique is $g$ such that $E=E(g)$? Is there a formula for $g$ such that $E=E(g)$ (see the note below for what kind of formula I envisage)? If $T\in SL(n,\mathbb{R})$ satisfies $E\_1 = T(E\_2)$ for two ellipsoids $E\_1$ and $E\_2$ of the same volume, does it follow that $T^t g\_1 T = g\_2$? What if $T$ is orthogonal or symplectic? Thanks for giving a hint or a reference.


Note: Given an ellipsoid $E$, then, I guess, $A\_{ij} = \int\_E (r^2 \delta\_{ij} - r\_i r\_j) dV$ and $B\_{ij} = \int\_E r\_i r\_j d V$ are inner products; they don't give $g$, though, but are related to it. It is clear from these formulas that if $E\_1 = T(E\_2)$, then $T^t A\_1 T = A\_2$ provided $T\in SO(n)$, and $T^t B\_1 T = B\_2$ provided $T\in SL(n)$.



Ellipsoids and their defining inner product
The *Binet-Legendre metric* of an ellipsoid $E\subset \mathbb{R}^n$ is defined as $g\_F$, the metric dual to 
$$
g\_F^\*(\xi,\eta)=\frac{n+2}{\operatorname{Vol}(E)}\int\_E \xi(x)\eta(x) dx.
$$
where the volume and integral are computed using a translation invariant Lebesgue measure. Note that rescaling the choice of measure has no effect.
Vladimir S. Matveev, Marc Troyanov, *The Binet-Legendre Metric in Finsler Geometry*, arXiv:1104.1647 prove that $E=\{x\in \mathbb{R}^n|g\_F(x,x)\le 1\}$ for any ellipsoid $E$.
This is not quite an explicit integral formula, because you still need to invert the symmetric matrix $g\_F^\*$, in any linear coordinate system, to get $g\_F$, but matrix inversion is an explicit algebraic map.


