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Let $A,B$ be two finite subgroups of $SO(4)$ such that $A$ and $B$ are isomorphic as abstract groups. Can we find a $g \in SO(4)$ such that $$ gAg^{-1}=B? $$ If it is the case, does the same conclusion hold for $SO(n)$?

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$\newcommand{\mat}[1]{\left(\begin{matrix}#1\end{matrix}\right)}$ Identifying $SO(4)$ with the group of deteminant-one matrices $Q\in \mathrm{M}_4(\mathbb R)$ such that $Q^TQ=QQ^T=\mathrm{Id}$, take

$$A=\lbrace\mathrm{Id},\mat{-1\\&-1\\&&-1\\&&&-1}\rbrace\quad\text{and}\quad B=\lbrace\mathrm{Id},\mat{1\\&-1\\&&1\\&&&-1}\rbrace.$$

Then both are subgroups of order $2$, and they cannot be conjugate, since their non-trivial elements have different characteristic polynomials.

Edited: The group $A$ was changed following a comment by TastyRomeo, the original group was a subgroup of $O(4)$, rather than of $SO(4)$.

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