Since I do not have enough reputations I could not comment directly at a similar question asked at https://mathoverflow.net/questions/219064/choosing-a-group-action-to-do-git-of-hypersurfaces?newreg=e451eed220d945dcb8711cd31ef335be.
According to Mumford's GIT, there is a linearization of the line bundle $\mathscr{O}(n)$ for the standard action of $\text{PGL}(n)$ on $\mathbf{P}_{k}^{n-1}$. I would like to see an explicit description of this action on the global section of $\mathscr{O}(n)$, which is a representation of the projective linear group $\text{PGL}(n)$. I tried the case $n=2$ and the representation of $\text{GL}(2)$ on $\text{Sym}^{2}(k^{2})$ coming from the standard representation $\text{id}:\text{GL}(2)\rightarrow\text{GL}(2)$. However, an explicit computation suggests that a matrix $\begin{pmatrix} a& b \\ c& d \end{pmatrix}$ is sent to $\begin{pmatrix} a^2 & ab & b^2\\ 2ac& ad+bc & 2bd\\ c^2 & cd & d^2 \end{pmatrix}$ under the homomorphism $\text{GL}(2)\rightarrow \text{GL}(\text{Sym}^2 k^2)\simeq\text{GL}(3)$. It does not appear that the kernel of this homomorphism is the center.
Since Jason Starr mentioned that the action of $\text{PGL}(n)$ lifts to an action on the space of degree $d$ homogeneous polynomials as long as $n$ divides $d$, I was expecting the representation of $\text{PG}L(n)$ on the global sections of $\mathscr{O}(n)$ comes from the symmetrized representation of the general linear group. Nevertheless, the above calculation suggests otherwise. Maybe I did something wrong or maybe the linearization is abstract?
To get an action of $\mathrm{PGL}(n)$, you should twist the $\mathrm{GL}(n)$ action by (a power of) the determinant. For instance, in the example with the action of $\mathrm{PGL}(2)$ on $\mathrm{Sym}^2(\Bbbk^2)$, you should divide your 3-by-3 matrix by $ad - bc$ (and by the way, I think the middle row of your matrix should be divided by 2).