$\newcommand{\euc}{\mathscr I}\newcommand{\R}{\mathbf R}$ Let $\euc(n)$ denote the the Euclidean group $\R^n\rtimes O_n(\R)$.
Recall that $\euc(n)$ acts on $\R^n$ as $(\mathbf x, T)\cdot \mathbf y=\mathbf x+T\mathbf y$ for all $(\mathbf x, T)\in \euc(n)$ and $\mathbf y\in \R^n$.
Consider the action of $\euc(n)$ on $(\R^n)^v$ defined as follows: $$ A\cdot(\mathbf y_1, \ldots, \mathbf y_v)=(A\cdot\mathbf y_1,\ldots, A\cdot\mathbf y_v) $$ for all $A\in \euc(n)$ and $(\mathbf y_1,\ldots, \mathbf y_v)\in (\R^n)^v$.
Fix a point $\mathbf p=(\mathbf p_1, \ldots, \mathbf p_v)\in (\R^n)^v$ and define a map $F:\euc(n)\to (\R^n)^v$ as $$ F(A)=(A\cdot\mathbf p_1,\ldots,A\cdot\mathbf p_v) $$ for all $A\in \euc(n)$.
Since $F^{-1}(\mathbf p)$ is an isotropy subgroup of a Lie group action, we know that $F^{-1}(\mathbf p)$ is a Lie subgroup of $\euc(n)$. Now for some reason unknown to me, we can give a manifold structure to $\euc(n)/F^{-1}(\mathbf p)$ such that the natural projection map $\pi:\euc(n)\to \euc(n)/F^{-1}(\mathbf p)$ is a smooth submersion.
For set theoretical reasons, we can get a map $\bar F:\euc(n)/F^{-1}(\mathbf p)\to (\R^{n})^v$ such that $F=\bar F\circ \pi$. Since $\pi$ is a smooth submersion, we infer that $\bar F$ is a smooth map.
My Question: Now on pg. 283 of this paper (titled 'The Rigidity of Graphs' and written by B. Roth and L. Asimow), it is said that the map $\bar F:\euc(n)/F^{-1}(\mathbf p)\to \text{im}(\bar F)$ is a diffeomorphism.
It is not clear to me why is this map a diffeomorphism. In fact, it is not even clear why is $\text{im}(\bar F)$ a smooth manifold.