Since you may ask about $\mathbb{RP}^n$, I may point out that you can embed $\mathbb{RP}^n$ into $\mathbb{R}^{(n+2)(n+1)/2-1}$ with a transitive action of $O(n+1)$.

From e.g. Exercise 5-C of Milnor-Stasheff, $\mathbb{RP}^n$ is the space of symmetric idempotent $(n+1)\times (n+1)$ matrices with trace $1$. This lives inside $\mathbb{R}^{(n+2)(n+1)/2-1}$ as the space of symmetric $(n+1)\times (n+1)$ matrices with trace $1$. $O(n+1)$ acts by conjugation on this space as isometries, and preserving the subspace of idempotent matrices.

An exceptional case is $\mathbb{RP}^3 \cong SO(3)$. The group $SO(3) \subset \mathbb{R}^9$ as $3\times 3$ matrices. But we can do a bit better: we may think of $SO(3)$ as pairs of orthogonal unit vectors $(v_1,v_2)\in (\mathbb{R}^3)^2$. This gives an embedding of $SO(3)\subset S^5\subset \mathbb{R}^6$ with a transitive group action. I believe that this special embedding exists since $so(4)=so(3)\oplus so(3)$.

One might be able to construct similar smaller dimensional embeddings using fibrations $S^1\to \mathbb{RP}^{2n+1}\to \mathbb{CP}^n$ and $\mathbb{RP}^3\to \mathbb{RP}^{4n+3}\to \mathbb{HP}^n$, but I haven't checked if they give smaller embeddings with isometric actions.

However, $\mathbb{RP}^{2n}$ is not a fibration (for the same reason that $S^{2n}$ is not a fibration). Hence, if we have an embedding $\mathbb{RP}^{2n}\subset \mathbb{R}^k$ and a transitive action by isometries $G\leq O(k)$, then the representation of the compact group $G$ must be irreducible. If not, then there is a splitting $\mathbb{R}^k=\mathbb{R}^{k_1}\times \mathbb{R}^{k_2}$ which is invariant under $G$. In this case, we get $v=(v_1,v_2)\in \mathbb{RP}^n\subset \mathbb{R}^k$, $v_i\in \mathbb{R}^{k_i}$, and $\mathbb{RP}^n = G\cdot v \to G\cdot v_1$. Hence we have a fibration $\mathbb{RP}^{2n} \to G\cdot v_1$, a contradiction unless $G\cdot v_1=\mathbb{RP}^{2n}$, in which case $k$ was not minimal.