The function $ f : \mathbb C^{2\times1} \rightarrow \mathbb C^{2\times1}$ is a conjugate linear transformation and has a coordinate matrix $ <\mathbf E^{*}, f(\mathbf E)>=E_2$ in respect to standard basis $ \mathbf E$ of $\mathbb C^{2\times1}$. Further, a basis $B=(b_1, b_2)$ (of $\mathbb C^{2\times1}$) is given, with:
$$ b_1= \begin{pmatrix} 0\\ -i \\ \end{pmatrix} $$
$$ b_2= \begin{pmatrix} -i\\ 1+i \\ \end{pmatrix} $$
We are supposed to calculate the coordinate matrix of the function in respect to $B$ and to verify if a composite function $f\circ f=f^2$ is also a conjugate linear transformation, or a simple linear one?
I thought of multiplying the coordinate matrix of a function $f$ with itself in order to obtain a matrix of $f\circ f$ and then to multiply this what I got with a generic vector in $\mathbb C^{2\times1}$ such as: $$ b_2= \begin{pmatrix} a+bi\\ c+di \\ \end{pmatrix} $$
to see if i get the conjugate complex function, but I don't know if that's the answer I'm looking for.
You can check directly how the composition $f \circ f$ behaves. If $v \in \mathbb{C}^2$ and $a \in \mathbb{C}$ then
$$ (f \circ f)(a v) = f(f(av)) = f(\overline{a} f(v)) = \overline{\overline{a}} (f(f(v))) = a ((f \circ f)(v)). $$
This shows that the composition of an anti-linear map with itself is linear.