Is this subset of $PSL(n,\mathbb{R})$ Zariski-closed? For some non-identity element $[A]\in PSL(n,\mathbb{R})$ ($[A]$ being the class of $A\in SL(n,\mathbb{R})$) and linearly independent vectors $x,y\in\mathbb{R}^n$, let $[x],[y]$ denote the classes of $x$ and $y$ in the projective space $\mathbb{R}P^{n-1}$. We then define
$$X=\{s\in PSL(n,\mathbb{R})\,|\,s[A]s^{-1}[x]=[y]\}.$$
My question is: 

Is $X$ Zariski-closed in $PSL(n,\mathbb{R})$?

Clearly its pre-image in $SL(n,\mathbb{R})$ is
$$\{B\in SL(n,\mathbb{R})\,|\,BAB^{-1}x\in\mathbb{R}^\times y\},$$
but that subset does not look Zariski-closed to me. (I really need it to be.) I have a feeling that it might boil down to the subset $$\{B\in SL(n,\mathbb{R})\,|\,BAB^{-1}x=y\},$$ which is clearly the zero set of $n$ real $n^2$-variable polynomials, but I don't see how.

Consider the polynomial map $f : \operatorname{SL}(n,\mathbb{R}) \to \wedge^2(\mathbb{R}^n)$ defined by $f(B) = (BAB^{-1}x) \wedge y$. Observe that since $Cx \neq 0$ for all $C \in \operatorname{SL}(n,\mathbb{R})$, we have $f(B) = 0$ if and only if $BAB^{-1}x \in \mathbb{R}^\times y$. Therefore, the pre-image of $X$ in $\operatorname{SL}(n,\mathbb{R})$ is the vanishing set of $\binom{n}{2}$ homogeneous polynomials of degree $n$ in the entries of $B$. Hence, $X$ is Zariski-closed in $\operatorname{PSL}(n,\mathbb{R})$.
