Is there an embedding (i.e. injective continuous map) 


$$\phi:\Bbb R \Bbb P^2\hookrightarrow S^4\subseteq\Bbb R^5$$ 


of the projective plane $\Bbb R\Bbb P^2$ into the $4$-sphere, that is *transitive*, i.e. for any two $x,y\in \Bbb R\Bbb P^2$ there is an orthogonal transformation $T\in\mathrm{O}(\Bbb R^5)$ that fixed the image $\mathrm{im}(\phi)$ set-wise, and has $Tx=y$?


Is $\Bbb R^5$ the lowest dimensional space in which such an embedding is possible, or do we need even more dimensions?


I may ask the same question for $\Bbb R\Bbb P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.



Transitive embedding of the projective plane $\Bbb R P^2$ into the $4$-sphere
Yes. You take each vector $v \in \mathbb{R}^3$ to the vector $v \cdot v \in \operatorname{Sym}^2\mathbb{R}^3=\mathbb{R}^6$. This takes each unit vector $v$ to the same place as $-v$. So it descends to $S^2/\pm 1=\mathbb{RP}^2$. 
If we identify each element of $\operatorname{Sym}^2\mathbb{R}^3$ with a symmetric matrix by identifying $\sum a\_{ij} e\_i \cdot e\_j$ with $A=(a\_{ij})$, then the action of $\operatorname{SO}(3)$ is by conjugation on these symmetric matrices. The element $e\_1 \cdot e\_1 + e\_2 \cdot e\_2 + e\_3 \cdot e\_3$ is identified with the identity matrix. The action preserves identity matrix, so preserves a 1-dimensional subspace in $\mathbb{R}^6$. We quotient out that subspace to get an action of $\operatorname{SO}(3)$ on $\mathbb{R}^5$. This is well known in representation theory as the unique 5-dimensional irreducible representation of $\operatorname{SO}(3)$. The metric $\left<A,B\right>=\sum\_{ij} A\_{ij} B\_{ij}$ is clearly preserved on $\mathbb{R}^6$, and $\left<v\cdot v,v\cdot v\right>=1$ in this metric. You take the quotient metric on $\mathbb{R}^5$, i.e. project to the orthogonal complement of that 1-dimensional subspace.


