Let $G = PSL_d(\mathbb{R})$ and $K = SO(d)$. It is a known fact that the symmetric space $G/K$ can be identified with the space $X_d$ defined by the space of inner products in $\mathbb{R}^d$ up to homothety.
This identification can be done by considering the action of $SL_d(\mathbb{R})$ in $X_d$ given by $$g.o(v, w) = o(g^{-1}v, g^{-1}w)$$ which is transitive and has as its stabilizer conjugates of $SO(n)$.
What I don't understand is how to prove that every isometry of $X_d$ is an element of $PSL_d(\mathbb{R})$.
EDIT: Let $\pi_o(g) = g.o$, with $o \in X_d$. If you compute the derivative of the map $\pi_o$ you get a map $\pi^o: sl_d(\mathbb{R}) \to T_oX_d$ which is an isometry when restricted to the $o$ antisymmetric matrices. You cand then give a Riemannian metric in $X_d$ using this identification and the Killing form in $sl_d(\mathbb{R})$