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Let $\mathrm{SO}(3) = \mathrm{SO}(3,\mathbb{R})$. Suppose $\mathrm{SO}(3) \curvearrowright \mathbb{R}^5$ via conjugation on traceless symmetric $3\times 3$ real matrices: $$\begin{pmatrix} a & c & d \\ c & b & e \\ d & e & -(a+b) \end{pmatrix}. $$

The stabilizer of the vector $M = (a,b,c,d,e) = (1,1,0,0,0)$ can be seen to be $\mathrm{O}(2)$. Thus, $\mathrm{SO}(3)/ \mathrm{O}(2) \simeq \mathbb{R}\mathrm{P}^2$ is an embedded submanifold of $\mathbb{R}^5$. See for example: Are closed orbits of Lie group action embedded?.

Is there a way to explicitly parametrize $\mathbb{R}\mathrm{P}^2$ in $\mathbb{R}^5$ using the previous? That is, can I write $x_1 = f_1(\alpha,\beta), \dots, x_5 = f_5(\alpha,\beta)$ for some smooth functions $f_1,\dots,f_5$ and parameters $\alpha,\beta$? My thought was to write $R \in \mathrm{SO}(3)$ in the fundamental representation (Euler angles) and then compute $R M R^{-1}$ and associate the coordinates in the obvious way. However, this gives $x_1,\dots, x_5$ in terms of the three Euler angles, whereas we need them in terms of two parameters (since $\mathbb{R}\mathrm{P}^2$ has dimension $2$).

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The O(2) you mod out by is rotation in the XY plane. You can make it an Euler angle. Choose your Euler angle convention so that R_z is the last (or maybe first; I hate Euler angles), and then it should be ignorable on the submanifold stabilizing M.

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