By Nash embedding theorem, every smooth Riemann manifold $M$ admits an isometric embedding $i:M \rightarrow \Bbb{R}^n$ for some $n$. However, such an embedding seems to 'lose symmetry', i.e. the isometry of $M$ may not be seen 'directly' when we look into the image of $i$ surrounded by $\Bbb{R}^n$. So how symmetric the embedding could be, determined by the Riemann structure on $M$?
Let's describe it in a more mathematical way: Let $\operatorname{Iso}(M)$ be the isometric group of $M$. Given $g \in \operatorname{Iso}(M)$, let $i_*(g)=i \circ g \circ i^{-1}$ be the isometry induced by $g$ on the image of $i$. Let $\operatorname{Ex}(i)$ be the largest subgroup of $\operatorname{Iso}(M)$ (not necessarily unique) such that $\forall h \in \operatorname{Ex}(i)$, $i_*(h)$ can be extended to an isometry of the surrounding $\Bbb{R}^n$. Is there a way to determine $\operatorname{Ex}(i)$?
This is a big topic, so a specific question is asked here: If we already have $\operatorname{Iso}(M) \le E(m)$ for a certain $m$, where $E(m)$ denotes the $m$-dimensional Euclidean group, must there be an embedding $i:M \rightarrow \Bbb{R}^n$ for some $n \ge m$ such that $\operatorname{Ex}(i)=\operatorname{Iso}(M)$? The situations of $M= \Bbb{R} \Bbb{P}^2$ and $M=L(p,q)$ (the lens space) endowed with the constantly curved Riemann structure are interested.