How do you show that conjugate mapping, $ f(z)=\bar {z}$ isn't linear? let $x_1= a+ib,x_2= c+id,k=$scalar

$f(x_1,x_2)=f(x_1) + f(x_2)$

$f(a+ib + c + id)=(a+c)-i(b+d)$

$f(a+ib)+f(c+id)=(a+c) - i(b+d)$

$f(kx_1)=kf(x_1)$

$f(k(a+ib))= k(a-ib)$

$kf(x_1)=k(a-ib)$

Looks linear to me, but tutor said it wasn't. please help.
There is a distinction between $\mathbb{R}$-linear and $\mathbb{C}$-linear.

So, here, for instance, if we take $w$ to be a scalar in $\mathbb{C}$, then

$$f(wz) = \bar{w} f(z).$$

Now, if $\Im w = 0$, then this is the required relation for linearity, but if $\Im w \neq 0$, then we don't have linearity. Therefore, $f$ is $\mathbb{R}$-linear but not $\mathbb{C}$-linear.

Regarding field automorphisms. Lets say we have fields $K$ and $L$ and $L\subseteq K$. If $\phi:K \to K$ is a field automorphism and $\phi(L) = L$, then $\phi$ is always $L$ linear, but only $K$ linear if it is the identity map.
$T(z) = \overline{z}$

$T(wz) = (w\overline{z}) = (\overline{w}\overline{z}) = \overline{w}f(z) \ne wf(z)$

Therefore not linear (consider the bar on the right of w and z as it is on the upper).