Suppose we have a function $g :\mathbb{R}^n \to \mathbb{R}$ such that for every $x \in \mathbb{R}^n$ $$ g(Ax)= g(x), \tag{$*$} $$ for all orthogonal matricies $A$. Then $g(x)$ is only a function of $\|x\|$.
In other words for all $x, y$ such that $x^Tx=y^Ty$ we have that $f(y)=f(x)$. That is the function has the same value on spheres of the same radius.
My question: Suppose a function $f:\mathbb{R}^n \to \mathbb{R}$ has the same value on ellipses of the same radius can we find a condition similar to that in $(*)$ that characterizes this property of $f$.
More concretly, suppose that for some given symetric positive definite matrix $B$ the function $f$ satifies $$ f(x)=f(y), \ \forall x^TBx=y^TBy. \tag{$**$} $$ Can we find a condition similar or akin to the one in $(*)$ to test whether the function $f$ satisfies $(**)$.
Note that $x^TBx=c$ defines an ellipse in $\mathbb{R}^n$.
There are infinitely many ways to find a $C$ with $C^T C = B.$ One of these ways is called the Cholesky decomposition, so I use the letter $C.$
Now, we have $B = C^T C.$ Suppose we take an orthogonal matrix $A^T A = I.$ Next, for convenience of typing, define $$ D = C^{-1}. $$ Finally, define $$ P = DAC. $$ We calculate $$ P^T BP = C^T A^T D^T C^TC DAC = C^T A^T I^T I A C = C^T A^T A C = C^T I C = C^T C = B $$
Next, given some $Q$ with $Q^T BQ = B,$ can you use $C,D$ to construct an orthogonal matrix using $Q \; ?$ You might try assuming that $Q = D WC$ with $W$ orthogonal and solve for $W,$ then check that $W$ really is orthogonal the way I did.