Adjoint representation of $Iso(\mathbb{R}^n)$ on $\mathfrak{iso}(\mathbb{R}^n)$

Question

It is known that the isometry group of $\mathbb{R}^n$ is a semidirect product of $O(n)$ (giving rotations) and $\mathbb{R}^n$ (giving translations). I am trying to see given $g \in Iso(\mathbb{R}^n)$, if there is a convenient formula for $Ad_g (X)$, where $X \in \mathfrak{iso}(\mathbb{R}^n)$. When $g \in O(n)$, I can see that $Ad_g(X) = g Xg^{-1}$. I am not sure what happens when $g \in \mathbb{R}^n$. Any suggestion is helpful, thanks!

Answer 1

It might be helpful to note that the group $Iso(\mathbb R^n)$ has a faithful representation in $GL(n+1,\mathbb R)$: $$ (b,A) \mapsto \left( \begin{matrix} A & b\\ 0 & 1 \end{matrix} \right), $$ where $A\in O(n)$ is a rotation/reflection and $b\in \mathbb R^n$ is a translation. For any matrix group, the formula $Ad_g(X) = gXg^{-1}$ still holds.