Basically what the title says. Let $f$ and $g$ be two polynomials with real coefficients of degree $m$, in $n$ real variables. The solutions of the equation $f(x_1,\ldots,x_n) = 0$ then define a hypersurface in $\mathbb{R}^{n}$. We can now apply a similarity transformation $A$ to each tuple $x\ = (x_1,\ldots,x_n)$ and obtain a new hypersurface consisting of the points $Ax$. Now let $g$ be the polynomial that has precisely $Ax$ as its zero set. How do you now obtain $g$ from $f$?
My understanding so far is that since $A$ is a similarity transformation, it must be a scalar multiple of an orthogonal transformation. For simplicity, we can assume the scalar to be equal to one. There now needs to be a way to translate this $\text{O}(n)$ action on $\mathbb{R}^n$ to some action on the polynomial ring containing $f$ and $g$ which maps $f$ onto $g$, but i have no idea what this action would look like.
For some context: This question arises from physics, where the surface of the eight-dimensional state space of a three-level quantum system can be described by a polynomial equation of degree 3, in 8 variables. I am currently studying 3 and 4 dimensional sections of this space, and there are some sections which have the same shape, despite being described by different equations. Especially, the equations
$3(x^2+y^2+z^2)-6\sqrt{3}xyz = 0$
and
$3(x^2+y^2+z^2)-3\sqrt{3}x(y^2-z^2) = 0$
both describe the same shape, according to the authors of this paper. However, the only similarity between these two equations i could find was that the sum of the absolute values of the coefficients is the same for both of them.
If $f,g$ are polynomials with $f(x)=0$ equivalent to $g(Ax)=0$ for an invertible transformation $A$, then $g=f(A^{-1}x)$.
In your case, the substitution $y\mapsto \frac{y-z}{\sqrt{2}}, z\mapsto \frac{y+z}{\sqrt{2}}$ takes the first equation to the second.