$U(2)$ fixes Orthogonal Complex Structures on $\mathbb{R}^4$? An orthogonal (almost) complex structure on a manifold M equipped with a metric g is a map $$J : TM\mapsto TM , J^2=-1$$ That preserves orientation and satisfies $g(Ju,Jv)=g(u,v)$
By identifying $\mathbb{R}^4$ with the pure imaginary unit Hamiltonian's we can see that for $x \in \mathbb{R}i+ \mathbb{R}j+ \mathbb{R}k $  s.t. $||x||=1$ , $x^2=-1$ hence each point of $S^2$ gives rise to an orthogonal complex structure. If we have such a structure, conjugating by elements of $SO(4)$ gives us all other orthogonal complex structures which I believe can be seen by $SO(4) \cong \frac{SU(2)\times SU(2)}{\mathbb{Z}_2}$ and identifying $SU(2)$ with the quaternions which can be used to generate all 4D rotations.
I'm reading through a paper by Atiyah on Twistor Theory and the paper states that given a complex structure of this type, it is fixed under the action of $U(2)$ . I think it's possible to see how $U(1)$ fixes such a structure by picking a complex structure  $u = xi+yj+zk$ with $x^2+y^2+z^2=1$ and then associating the unit quaternion $t = Cos(\theta) + u Sin(\theta)$ with it. Then $t^{-1}ut = u$ hence we have a free parameter $\theta$ that can be identified with $U(1)$ that fixes $u$.
However I was wondering how to see that $U(2)$ fixes an orthogonal complex structure on $\mathbb{R}^4$?
 A: Given $x $ as in your post, I'll write $J_x$ to denote the complex structure given by left multiplication by $x$.
Consider all elements of the form $(e^{tx}, q)$ with $t\in \mathbb{R}$ and $q\in Sp(1)$ (where $e^{tx}$ is shorthand for $\cos(t) + x\sin(t)$).  This is obviously a subgroup of $Sp(1)\times Sp(1)$, and I'll denote it by $H$.  The group $H$ is isomorphic to $S^1\times Sp(1)\cong S^1\times SU(2)$.
Note that the induced action by $H$ on $\mathbb{R}^4\cong \mathbb{H}$ fixes $J_x$ Indeed, since real numbers and $x$ both commute with $x$, we find \begin{align*}(e^{tx},q)\ast J_x(v) &= e^{tx} x v q^{-1}\\ &= x\left(e^{tx}v q^{-1}\right)\\ &= J_x( (e^{tx},q)\ast v).\end{align*}
We claim that the kernel of this action is $\langle(-1,-1)\rangle\in H$.  To see this, simply note that the kernel of the projection $Sp(1)\times Sp(1)\rightarrow SO(4)$ is precisely $\langle (-1,-1)\rangle$, and that $(-1,-1)= (e^{\pi x}, -1)\in H$.
Thus, we really have $H/\langle (-1,-1)$ acting on $\mathbb{R}^4$ and preserving $J_x$.  We need only show that $H/\langle (-1,-1)\cong U(2)$.
To that end, note that under any isomorphism $Sp(1)\rightarrow SU(2)$, $-1$ must map to $-I$.  So, we may view this as $S^1\times SU(2)/ \langle (-1,-I)\rangle$.  The later is well known to be isomorphic to $U(2)$, but for concreteness, one can easily verify that map $f:S^1\times SU(2)\rightarrow U(2)$ given by $f(z,A) = zA$ is surjective with kernel $\langle (-1,-I)\rangle.$
