Suppose that $T : \mathbb C^2 \to \mathbb C^2$ is an $\mathbb R$ linear map. $T(1,0) = (1,0)$, $T(0,1)=(0,1)$ and $T$ maps the $\mathbb C$-subspaces of $\mathbb C^2$ to $\mathbb C$-subspaces of $\mathbb C^2$. What is $T$?
My attempt: First I can show that $T$ is a surjective $R$ linear map, thus a bijective $R$ linear map. And I know that $T$ maps $(i,0)$ to $(k,0)$ where $k$ is a nonzero complex number, but I guess $k$ should either be $(i,0)$ or $(-i,0)$. I do not know how to prove it.
$C^2$ is a $4$-dimensional real vector space with basis $(1,0), (i,0), (0,1), (0,i)$ $T(1,0)C=(1,0)C$ you have $T(i,0)=(a,0)$ $a\in C$, you also have $T(0,i)=(0,b), b\in C$.
You have $T(1,1)=(1,1)$ so $T((1,1)C)=(1,1)C$, so you have $T(i,i)=(c,c)=T(i,0)+T(0,i)=(a,b)$ it results $a=b=c$.
$T(i,1)=T(i,0)+T(0,1)=(a,1)$, $T(-1,i)=T(-1,0)+T(0,i)=(-1,a)$ since $i(i,1)=(-1,i)$ we deduce that there exists a complex number $d$ such that $d(a,1)=(-1,a)$ $da=-1, d=a$ and $a^2=-1$ this implies $a=i$ or $a=-i$.
As you say, there is some nonzero $k$ such that $T(i,0)=(k,0)$, and there is similarly a nonzero $\ell$ such that $T(0,i)=(0,\ell)$. But so far you have only used a few of the complex subspaces; there are a lot more you can use. For any $z=x+iy\in\mathbb{C}$, you can consider the complex span of $(1,z)$; $T$ must send this to a complex subspace. Since $T(1,z)=(1,x+\ell y)$, this means that $T(i,iz)$ must be a complex multiple of $(1,x+\ell y)$. See what you can learn from this.
The details of how to finish from here are hidden below.
Since $T(i,iz)=T(i,-y+ix)=(k,-y+\ell x),$ this is a complex multiple of $(1,\ell)$ iff $-y+\ell x=k(x+\ell y)$. Since this must hold for arbitrary $x,y\in \mathbb{R}$, we must have $\ell=k$ and $k\ell=-1$. That is, either $k=\ell=i$ or $k=\ell=-i$. In the first case $T$ is the identity, and in the second case $T$ is complex conjugation on each coordinate. You can check that both of these do preserve complex subspaces, so both are possible.