# Are two isomorphic finite subgroups of $SO(4)$ conjugate?

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 $$SO(n)$$?

$$\newcommand{\mat}[1]{\left(\begin{matrix}#1\end{matrix}\right)}$$ 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 $$SO(4)$$.