a basic question about complex integrals Let
$$
\gamma_1:[a,b]\to\mathbb C,\qquad
\gamma_2:[a,b]\to\mathbb C
$$
be two one-to-one, continuously differentiable parametrizations of the same (oriented) curve $\Gamma$. In addition, suppose that $f(z)$ is a complex function which is continuous on a region containing $\Gamma$.
I know that if there exists a strictly increasing and continuously differentiable function $\alpha:[a,b]\to[a,b]$ such that
$\gamma_2(t)=\gamma_1(\alpha(t))$ for every $t\in[a,b]$, then
$$
(\ast)\qquad \int_{\gamma_1}f(z)dz=\int_{\gamma_2}f(z)dz.
$$
My question: Can we deduce $(\ast)$ merely by the fact that $\gamma_1,\gamma_2$ are parametrizations of the same curve? Is the existence of $\alpha$ not guaranteed?

If such $\alpha$ exists, then necessarily, $\alpha = \gamma_1^{-1} \circ \gamma_2$. The question is whether $\alpha$ is $C^1$. Counterexamples show that this is not true under the hypotheses you stated.
To state the result, I will change the domains of $\gamma_1, \gamma_2$ to be $(a_1, b_1)$ and $(a_2, b_2)$ respectively. The result is that if $\gamma_1$ and $\gamma_2$ are homeomorphisms onto their images $U_1, U_2 \subset \Gamma$ (which can be different) and are $C^1$ with nonvanishing derivative at each point, then the transition map $\gamma_1^{-1} \circ \gamma_2 : \gamma_2^{-1}(U_1 \cap U_2) \to \gamma_1^{-1}(U_1 \cap U_2)$ is $C^1$ with $C^1$ inverse. See discussion on pages 135-137 here for the proof of the $m$-dimensional version of the above result: https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/analmv.pdf
