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Given a surface $V(f) \subset \mathbb{P}^n$ for a homogeneous polynomial $f$ of degree $d$ on $\mathbb{P}^n$ and a linear transformation $g \in SL(n+1)$. Is the degree of the Hessian $H_f = V(\det (\frac{\partial f}{\partial x_i\partial x_j}))$ of $f$ equal to the degree of the Hessian $H_{f\circ g} = V(\det (\frac{\partial (f\circ g)}{\partial x_i\partial x_j}))$ of $f\circ g$? If so, why? If not, are there any restrictions under which this holds?

Many thanks in advance!

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If $f(x_0,...,x_n)$ is a homogeneous polynomial of degree $d$, its second partial derivatives $\frac{\partial ^2f}{\partial x_i\partial x_j}$ have degree $d-2$, so that the Hessian determinant $Hess(f)=det(\frac{\partial^2 f}{\partial x_i\partial x_j})$ is homogeneous of degree $(d-2)(n+1)$ or is zero.
Since $f\circ g$ is also homogeneous of degree $d$, we see that $Hess(f\circ g)$ is homogeneous of degree $(d-2)(n+1)$ or is zero.

Conclusion
Your varieties $H_f=V(Hess(f))$ and $H_{f\circ g}=V(Hess(f\circ g))$ are equal to $\mathbb P^n$ or are hypersurfaces of degree $(d-2)(n+1)$.