In Peter Olver's Complex Analysis book, it is stated in theorem 7.12 that complex integrals are invariant under conformal maps, and the proof is left as an exercise using change of variables, but I am having trouble reproducing it.
Claim: Let $\zeta = g(z)$ be a conformal map and assume $g$ maps the curve $C=\{z(t)\}$ to the curve $\Gamma=\{\zeta(t)=g(z(t))\}$. Let $F(\zeta)$ be related to $f(z)=F(g(z))$. Then
$$ \int_C f(z)\,dz=\int_\Gamma F(\zeta)\,d\zeta .$$
Attempted proof: $$\int_\Gamma F(\zeta)\,d\zeta = \int_{t_0}^{t_1} F(\zeta(t))\tfrac{d\zeta}{dt}\,dt = \int_{t_0}^{t_1} F(g(z(t)))\tfrac{d(g(z(t))}{dt}\,dt$$ $$ = \int_{t_0}^{t_1} F(g(z(t)))g'(z(t))\tfrac{dz}{dt}\,dt = \int_C f(z)g'(z)dz$$ and the last term is generally not equal to $\int_C f(z)\,dz$. Is there an incorrect step?
The question is in correct understanding of invariance here. Applying your reasoning for toy-model: $F(\zeta)=f(z)\equiv 1$, $\zeta=g(z)=2z$, $C=[0,1]\subset\mathbb{R}$ can be claryfying. Dou You expect $\int_0^2 1\, d\zeta$ to be equal to $\int_0^1 1\,dz$ or $\int_0^1 1\cdot 2dz$ in order to integral to be invariant?