This is closely related to a previous question:
Scale invariant definition of the Sobolev norm $\|\|_{m,\Omega}$ for $H^m(\Omega)$
This question focuses on the direct calculation (by change of variables) of the simplest one-dimensional case.
Let $\Omega\subset\mathbb{R}^n$ be a nonempty bounded open set. Denote the Lebesgue measure of $\Omega$ as $|\Omega|$. For $f\in H^m(\Omega)$, define
$$ \|f\|_{m,\Omega}^2=\sum_{|\alpha|\leq m}\color{blue}{\big|\Omega\big|^{\dfrac{2(|\alpha|-m)}{n}}} \int_{\Omega}|D^\alpha f|^2\ dx. $$ According the so-called "dimensional analysis" (see the comment to a previous question), the quantity
$$ C(\Omega):=\dfrac{1}{\big|\Omega\big|^{\dfrac{n-2m}{n}}}\cdot \|f\|_{m,\Omega}^2 $$ is scale invariant, namely, $C(\Omega)=C(\delta\Omega)$ for any $\delta>0$ where $$ \delta\Omega:=\{\delta x\mid x\in \Omega\}. $$
Note that we can rewrite $C(\Omega)$ as $$ C(\Omega)=\sum_{|\alpha|\leq m}\color{blue}{\big|\Omega\big|^{\dfrac{2|\alpha|-n}{n}}} \int_{\Omega}|D^\alpha f|^2\ dx \tag{*} $$
I would like test the statement above by a direct calculation of $C(\delta\Omega)$ using change of variables with the simple case $n=1$ and $\Omega=(0,1)$. In this case, $(*)$ becomes $$ C(\delta\Omega)=\sum_{k\leq m}\color{blue}{\delta^{2k-1}} \int_{\delta\Omega}|D^kf(x)|^2\ dx. $$ To show $C(\Omega)=C(\delta\Omega)$, it suffices to show that $$ \int_0^\delta|D^k f|^2\ dx=\delta^{1-2k}\int_0^1|D^k f(x)|^2\ dx. $$
But even in the case $k=1$, one can check that the identity $$ \int_0^\delta |f'(x)|^2\ dx=\frac{1}{\delta}\int_0^1|f'(x)|^2\ dx $$ is wrong.
What is going wrong here?
For the calculation of $C(\delta\Omega)$, when the domain changes from $\Omega$ to $\delta\Omega$, $f$ should changes to be $T_\delta f$ where $$ T_\delta f(y):=f(\frac{y}{\delta}),\quad y\in\delta\Omega. $$ It does not make sense at all to compare $\|f\|_{m,\Omega}$ with $\|f\|_{m,\delta\Omega}$ since if $f$ is defined on $\Omega$, it is not necessarily defined on $\delta\Omega$.
Instead, one should compare $\|f\|_{m,\Omega}$ with $\|T_\delta f\|_{m,\delta\Omega}$. Then everything makes sense with the change of variable formula. For instance, in the last identity in OP, one would instead get $$ \int_0^\delta |(T_\delta f)'(x)|^2\ dx=\frac{1}{\delta}\int_0^1|f'(x)|^2\ dx\tag{1} $$ Noting that by chain rules $$ (T_\delta f)'(x)=\frac{1}{\delta}f'(\frac{x}{\delta}) $$ we can check by change of variables that (1) is indeed true.