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

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

of the projective plane $\Bbb R P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in P^2\Bbb R$ 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 $\Bbb R P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.

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.

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

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

of the 2-dimensional projective space $\Bbb R P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in P^2\Bbb R$ 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 $\Bbb R P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.

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

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

of the projective plane $\Bbb R P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in P^2\Bbb R$ 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 $\Bbb R P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.

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

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

of the 2-dimensional projective space $P^2\Bbb R$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in P^2\Bbb R$ 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 $P^n\Bbb R$: what is the lowest dimensional Euclidean space needed for such an embedding.

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

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

of the 2-dimensional projective space $\Bbb R P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in P^2\Bbb R$ 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 $\Bbb R P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.