I'm trying to prove the Grassmann manifold $\mathrm G_k(\mathbb R^n)$ of $k$-dimensional linear subspaces of $\mathbb R^n$ is isotropic and symmetric.
By "isotropic" I specifically that for every point $P \in \mathrm G_k(\mathbb R^n)$, and for every pair of unit tangent vectors $u, v \in T_P \mathrm G_k(\mathbb R^n)$, there is an isometry $\phi : \mathrm G_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ fixing $P$ and with $d\phi_P u = v$. By "symmetric" I mean for every $P \in \mathrm G_k(\mathbb R^n)$, there is an isometry $\phi$ fixing $P$ and so that $d\phi_P = -\mathrm{Id} : T_P \mathrm G_k(\mathbb R^n) \to T_P \mathrm G_k(\mathbb R^n)$. There appear to be a few different definitions for each, and I want to be clear with which one I'm using.
My setting: $\mathrm G_k(\mathbb R^n)$ has a unique smooth structure with respect to which the natural action of $\mathrm{GL}(n,\mathbb R)$ on $\mathrm G_k(\mathbb R^n)$ is smooth. To define the Riemannian metric, we consider the Stiefel manifold $\mathrm V_k(\mathbb R^n)$ of $k$-tuples of orthonormal vectors in $\mathbb R^n$. This may be considered a submanifold of the space $\mathrm M(n \times k, \mathbb R)$ of $n \times k$ real matrices, with the Euclidean metric from identifying $\mathrm M(n \times k, \mathbb R) \approx \mathbb R^{nk}$. The map $\pi : \mathrm V_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ sending a $k$-tuple of orthonormal vectors to its span is a surjective smooth submersion. One can show $\mathrm O(k)$ acts on the right of $\mathrm V_k(\mathbb R^n)$ isometrically, vertically (meaning for $A \in O(k)$, $B \in \mathrm V_k(\mathbb R^n)$, we have $\pi(BA) = \pi(B)$), and transitively on fibers (meaning if $B$ and $B'$ are two orthonormal $k$-tuples with the same span, there's some $A \in \mathrm O(k)$ with $BA = B'$). It follows that there is a unique Riemannian metric on $\mathrm G_k(\mathbb R^n)$ with respect to which $\pi : \mathrm V_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ is a Riemannian submersion, meaning for each $B \in \mathrm V_k(\mathbb R^n)$, $d\pi_B : \ker\left(d\pi_B\right)^\perp \to T_{\pi(B)} \mathrm G_k(\mathbb R^n)$ is a linear isometry. Call this metric $g$ on $\mathrm G_k(\mathbb R^n)$.
My strategy: I know $\mathrm O(n)$ acts on the left transitively and isometrically on $\mathrm V_k(\mathbb R^n)$, and hence acts transitively and isometrically on $\mathrm G_k(\mathbb R^n)$, so $\mathrm G_k(\mathbb R^n)$ is homogeneous. So I only need to prove $\mathrm G_k(\mathbb R^n)$ is isotropic and symmetric at a single point; say the subspace $P = \mathbb R^k \subset \mathbb R^n$, spanned by the first $k$ coordinates. The isotropy group in $\mathrm O(n)$ of this point is $G_P := \mathrm O(k) \oplus \mathrm O(n-k)$. It seems intuitively obvious that with an appropriate choice of matrix $A \in \mathrm O(k) \oplus \mathrm O(n-k)$, the differential of the action map $\theta_A : \mathrm G_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ may either reflect the tangent space at $P$ or act transitively on unit tangent vectors, but I'm having trouble with the specifics. Plus the coordinates of the Grassmannian seem kind of weird and intimidating. Is there a coordinate-free way to make this argument rigorous?