Do Galilean (Euclidean) space transformations implies that time is absolute? I recently read a paper where it says "if space is universally Euclidean, then time is universal" and I don't understand some key points about the implication.
To put in context, the author argues, based on historical sources, that the name of Galilean transformations is misleading and it would be more appropriate to call them as Euclidean space-time transformations. Since in classical mechanics time is not a transformable quantity (like coordinates and velocities), an Euclidean space transformation for an event at points $(x,y,z)$ in the inertial frame $\mathcal{S}$ and measured in $(x',y',z')$ in $\mathcal{S}'$, where $\mathcal{S}'$ is another inertial frame that moves in the $+x$ direction at constant $v$ relative to $\mathcal{S}$, is given by
$$x'=x-vt;\quad y'=y;\quad z'=z \tag{1}$$
Then, as a mathematical consequence time is absolute. The "proof" started as:

A general time transformation equation is now added to Eq. (1). Then, without any assumption about the time transformation except linearity, the space-time transformation of an event measured as $x,y,z,t$ in $\mathcal{S}$, and $x',y',z',t'$ in $\mathcal{S}'$, can be written as
$$x'=x-vt;\quad y'=y;\quad z'=z;\quad t'=\alpha t-\beta x$$
where $\alpha$ and $\beta$ allow $t'$ to be a linear function of $t$ and $x$. Linearity of the transformation equations is necessary in order to guarantee any particular event in one frame appears as a single event, without echoes, in the other frame of reference.

After some steps, the author obtains that $t'=t$, as expected. What I don't understand is why we have to assume linearity. I don't see how this property guarantee that there are no "echoes" from a single event. In general, if I write time in $\mathcal{S}'$ as $t'=\alpha t^n+\beta x^m$, how can I know that $n=m=1$?
 A: Firstly, I'd like to mention that for the equation
$$t'=\alpha t-\beta x$$
to be dimensionally consistent $\alpha$ needs to be a dimensionless constant, but $\beta$ needs dimensions of inverse velocity (time / length). The transformation ought to depend on $v$, but we can't simply put $v$ in the denominator because that's undefined when $v=0$. One possible form of the equation with both $\alpha$ & $\beta$ as dimensionless constants is
$$t'=\alpha t-\beta vx/c^2$$
where $c$ is some constant speed, not necessarily the speed of light.

The reason that linearity is important is that linear equations are guaranteed to have, at most, one solution. And the inverse of a linear equation is also linear, so the inverse has (at most) one solution, too.
By symmetry, a coordinate transformation has to work both ways: the same equation that transforms from the unprimed frame to the primed frame can be used to transform from the primed frame to the unprimed frame, with only a minor change that takes into account the change of the sign of $v$. (If our axes are aligned, and in my frame I  observe you to be moving in the $+x$ direction, in your frame you observe me to be moving in the $-x$ direction).
If the coordinate transformation equation isn't linear, then it's possible for the equation & / or its inverse to have multiple solutions, i.e., the undesirable "echoes" mentioned in the paper.
A: If the time absolute then die equations of motion in S and S' space must be the equal.
because the potential energy ,$U=U(x,y,z)$, in S and S' is equal  ,we just have to calculate  the kinetic energy and if the kinetic energy in S and S' is the equal then the  EOM's are the equal. I assume that the mass is one.
I) Kinetic Energy  in S system
with the position vector
$$\vec{R}=\left[ \begin {array}{c} x\\y\\
z\end {array} \right]
$$
thus the the Kinetic energy
$$T=\frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right) \tag 1$$
II) Kinetic Energy  in S'  system
with $x'=x-v\,t$ you get the position vector
$$\vec{R}=\left[ \begin {array}{c} x-vt\\ y
\\ z\end {array} \right] 
$$
and
$$2\,T=\frac{\partial \vec{R}}{\partial \vec{q}}\,\vec{\dot{q}}+\frac{\partial \vec{R}}{\partial t}$$
with:
$$\vec{q}= \left[ \begin {array}{c} x\\ y\\
z\end {array} \right] 
$$
$\Rightarrow$
$$T=\frac 1 2\left({{\dot{x}}}^{2}-v{\dot{x}}\,+{v}^{2}+\,{{\dot{y}}}^{2}+\,{{
\dot{z}}}^{2}\right)
\tag 2$$
III) Kinetic Energy  in S'  system
with $t'=\alpha\,t+\beta\,x\quad \Rightarrow\quad,t=\frac{t'+\beta\,x}{\alpha}$
the position vector is now
$$\vec{R}=\left[ \begin {array}{c} x-{\frac {v \left( { t'}+\beta\,x
 \right) }{\alpha}}\\ y\\ z
\end {array} \right]
$$
and
$$2\,T=\frac{\partial \vec{R}}{\partial \vec{q}}\,\vec{\dot{q}}+\frac{\partial \vec{R}}{\partial t'}\frac{\partial t'}{\partial t}$$
$\Rightarrow$
$$T=\frac{1}{2}\left({\frac { \left( -\alpha+\beta\,v \right) ^{2}{{\dot{x}}}^{2}}{{\alpha}^
{2}}}+2\,{\frac {v \left( -\alpha+\beta\,v \right) {\dot{x}}}{\alpha}}+
{v}^{2}+{{\dot{y}}}^{2}+{{\dot{z}}}^{2}
\right)\tag 3$$
The equations of motion with the kinetic energy eq. (1) and eq. (2) are equal.
If you choose $\beta=\frac{2\alpha}{v}$ then the kinetic energy  eq. (2) will be equal to the kinetic energy eq. (3).
Thus:
if the EOM's are equal in all  three cases then the time must be absolute
