As rita said $\mathbb{P}(1,\dots,1,k)$ is naturally isomorphic to the cone in $\mathbb{P}^{N+1}$ over the $k$-th embedding (Take the map which sends $(x_1:\dots:x_n:y)$ onto $((x_1)^k:(x_1)^{k-1}x_2:...:(x_n)^k:y)$ where the $N+1$ first coordinates are the monomials of degree $k$ in $x_1,\dots,x_n$), so there is a natural morphism from the group $G=\mathrm{Aut}(\mathbb{P}(1,\dots,1,k))$ to $\mathrm{PGL}(n,\mathbb{K})$. However, the kernel is not the one which was described in the above answer.

We can in fact give $G$ more explicitly (because there are a priori many extensions given two groups):

We choose $k>1$ (otherwise the description is different and obvious). We identify $\mathbb{K}^{N+1}$ with the set of homogeneous polynomials of degree $k$ in $n$ variables. The group $\mathrm{GL}(n,\mathbb{K})$ naturally acts on $\mathbb{K}^{N+1}$.

Let $H$ be the semi-direct product $\mathbb{K}^{N+1}\rtimes \mathrm{GL}(n,\mathbb{K})$. There is a natural surjective map $H\to G$, that we describe now:

The action of $\mathbb{K}^{N+1}$ on $\mathbb{P}(1,\dots,1,k)$ is given by $(x_1:\dots:x_n:y)\mapsto (x_1:\dots:x_n:y+P(x_1,\dots,x_n))$ where $P\in\mathbb{K}^{N+1}$ is the corresponding polynomial.

The action of $\mathrm{GL}(n,\mathbb{K})$ on $\mathbb{P}(1,\dots,1,k)$ is given by the action on $x_1,\dots,x_n$.

It yields thus a morphism $H\to G$ whose kernel is the subgroup $L$ of $\mathrm{GL}(n,\mathbb{K})$ consisting of diagonal matrices of the form $\{\lambda I| \lambda^k=1\}$.

The group $G=\mathrm{Aut}(\mathbb{P}(1,\dots,1,k))$ is thus equal to the quotient of $\mathbb{K}^{N+1}\rtimes \mathrm{GL}(n,\mathbb{K})$ by the subgroup $L$.

The surjective morphism $G\to \mathrm{PGL}(n,\mathbb{K})$ corresponds to the projection on $\mathrm{GL}(n,\mathbb{K})/L$ followed by the quotient by the image of all diagonal matrices (we have first killed only finitely many and then kill all others). The kernel of this map is thus equal to $\mathbb{K}^{N+1}\rtimes\mathbb{K}^{*}/L$.