In the variational calculus, we discuss general case variation where endpoints vary as well. (i.e $x_0$ and $x_1$ vary as well). By using some calculations, we end up with:
$\delta J = \left[\frac{\partial F}{\partial y'}\delta y + (F - y'\frac{\partial F}{\partial y'})\delta x\right]_{x=x_0}^{x=x_1}$
Note that $\delta J$ is already a first order variation and it doesn't include euler lagrange, because assuming $y(x)$ is a true path, even doing the variation with fixed endpoints must cause $\delta J$ to be 0, so that euler lagrange part vanishes. Also, keep in mind that since $y(x)$ is a true path, $\delta J = 0$.
Question:
in the following book, author uses transformation(I won't include them here as it's short and can be seen on the link in a clearer way) and here is what's important. For sure, he replaces $\delta x$ and $\delta y$ by the transformations. Important thing though, is that he only makes $\delta J$ be 0 if and only if functionals are invariant.
My question is the following: Why can he only make $\delta J=0$ only in the case when functionals are invariant ? When you have a functional $J[y]$, you assume that $y(x)$ is a true path and then using transformation on the true path is the same thing as varying the path which should cause first order variation to be 0.