Why the zero-order term in a variational transformation of coordinates should be identically the same as the old coordinates? In the Ref.[1, page 61] the author proposes that transformations between two coordinate systems can be described by a continuous parameter $\varepsilon$ such that when $\varepsilon=0$ the original coordinates are recovered.
The mapping between these two systems imply the existence of functions $T$ and $Q$ such that
\begin{align}
t\rightarrow t^\prime &=T\big(t,q^\mu,\varepsilon\big),\tag{4.1.1}\label{eq1}\\
q^\mu\rightarrow {q^\prime}^\mu &=Q^\mu\big(t,q^\nu,\varepsilon\big).\tag{4.1.2}\label{eq2}
\end{align}
The author states that if $\varepsilon$ is sufficiently small, then we can to expand the functions $T$ and $Q$ in Taylor series about $\varepsilon=0$, such that:
\begin{align}
t^\prime &=t+\varepsilon\bigg(\frac{\partial T}{\partial\varepsilon}\bigg)_{\varepsilon=0}+O\big(\varepsilon^2\big),\tag{4.1.7}\label{eq7}\\
{q^\prime}^\mu &=q^\mu+\varepsilon\bigg(\frac{\partial Q^\mu}{\partial\varepsilon}\bigg)_{\varepsilon=0}+O\big(\varepsilon^2\big),\tag{4.1.8}\label{eq8} 
\end{align}
where the author the author identifies $t^\prime=t$ and ${q^\prime}^\mu=q^\mu$ when $\varepsilon=0$.
Also, according to the author, the coefficients of $\varepsilon$ to the first power are called the ''generators'' of the transformation and they can be denoted by
\begin{align}
\tau &\equiv\bigg(\frac{\partial T}{\partial\varepsilon}\bigg)_{\varepsilon=0}=\tau\big(t,q^\mu\big),\tag{4.1.9}\label{eq9}\\
\zeta^\mu &\equiv\bigg(\frac{\partial Q^\mu}{\partial\varepsilon}\bigg)_{\varepsilon=0}=\zeta^\mu\big(t,q^\mu\big).\tag{4.1.10}\label{eq10}
\end{align}
Now, let me introduce how I'm seeing the problem. The Taylor series expansion around $\varepsilon = 0$ are written as
\begin{align}
T\left(  t,q^{\mu},\varepsilon\right)    & =T\left(  t,q^{\mu},0\right)
+\left(  \dfrac{\partial T}{\partial\varepsilon}\right)  _{\varepsilon
=0}+O\left(  \varepsilon^{2}\right),  \\
Q^\mu\left(  t,q^{\mu},\varepsilon\right)    & =Q^\mu\left(  t,q^{\mu},0\right)
+\left(  \dfrac{\partial Q^\mu}{\partial\varepsilon}\right)  _{\varepsilon
=0}+O\left(  \varepsilon^{2}\right).
\end{align}
But, according to the author, on the left side, we have $t^\prime=T\big(t,q^\mu,\varepsilon\big)$ and ${q^\prime}^\mu=Q^\mu(t,qν,\varepsilon\big)$. On the other hand, on the right side, the zero-order term is $t=T\big(t,q^\mu,0\big)$ and $q^\mu=Q^\mu(t,q^\mu,0\big)$, while the first order term $\tau\big(t,q^\mu\big)=\big(\partial T/\partial\varepsilon\big)_{\varepsilon=0}$ and $\zeta^\mu\big(t,q^\nu\big)=\big(\partial Q^\mu/\partial\varepsilon\big)_{\varepsilon=0}$.
My question is centered on the zero-order term because I think that instead of being $T\big(t,q^\mu,0\big)=t$ and $Q^\mu\big(t,q^\nu,0\big)=q^\mu$ it should be $T\big(t,q^\mu,0\big)=\mathcal{T}\big(t,q^\mu\big)$ and $Q^\mu\big(t,q^\nu,0\big)=\mathcal{Q}^\mu\big(t,q^\nu\big)$ since we are searching for the most general form possible.
So, what should be the argument for making a more restrictive choice such as $T\big(t,q^\mu,0\big)=t$ and $Q^\mu\big(t,q^\nu,0\big)=q^\mu$?
In my opinion, Eqs. \eqref{eq7} and \eqref{eq8} should be written as
\begin{align}
t^\prime &=\mathcal{T}\big(t,q^\mu\big)+\varepsilon\tau\big(t,q^\mu\big)+O\big(\varepsilon^2\big),\tag{B1}\label{eqB1}\\
{q^\prime}^\mu &=\mathcal{Q}^\mu\big(t,q^\mu\big)+\varepsilon\zeta^\mu\big(t,q^\mu\big)+O\big(\varepsilon^2\big).\tag{B2}\label{eqB2} 
\end{align}
$^{[1]}$ Dwight E. Neuenschwander, Emmy Noether's Wonderful Theorem
 A: If you replace eqs. (4.1.1) and (4.1.2) by your (A) equations you actually kill the extra parameter introduced $\varepsilon$, rendering the rest of the chapter pointless. 
The "generator" term comes from group theory, and in page 64 you find a very brief note: "[...] Because a set of such transformations has an identity element and each transformation has an inverse, the transformations form a group [...]". 
And $t'=t$ holds in the limiting case $\varepsilon \rightarrow 0$, from where you can deduce the form of your function $\mathcal{T}$ not to include $q^\mu$, and $\mathcal{Q}$ not to include $t$.
Look at example in equation (4.1.3) with $\varepsilon$ being a rotation angle about the $z$ axis, how would a series expansion for $t'$ look like in that case?
A: After some reflection, I think I came to an understanding of what should be the answer to the problem. Transformations \eqref{eqB1} and \eqref{eqB2} should be viewed not only as transformations between coordinates but also as perturbations around the original coordinates such that at the boundary where $\varepsilon\rightarrow 0$ these must be retrieved to recover. In this sense, in order for such a criterion to be satisfied, we must choose (or configure) the functions $\mathcal{T}$ and $\mathcal{Q}^\mu$ as $\mathcal{T}\big(t,q^\mu\big) = t$ and $\mathcal{Q}^\mu\big(t,q^\mu\big) = q^\mu$.
Basically, I think this is the same answer that @fredwhileshavin wanted to give in the previous post, but in a veiled way.
Well, I think I'm on the right track for a good answer, but I leave it to the community for judgment.
