
$A. From Einstein's derivation of the Lorentz Transformation
Einstein proved first that the transformation between the inertial frames $\rm S_1,\rm S_2$ of the Figure(1) must be, except an undetermined factor $\phi\left(\boldsymbol{\upsilon}\right)$
\begin{align}
x_2 &\boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)\gamma_\upsilon \left(x_1\boldsymbol{-}\upsilon t_1\right)
\tag{01a}\label{01a}\\
y_2 & \boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)y_1
\tag{01b}\label{01b}\\
z_2 & \boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)z_1
\tag{01c}\label{01c}\\
t_2 &\boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)\gamma_\upsilon \left(t_1\boldsymbol{-}\dfrac{\upsilon}{c^2}x_1\right)
\tag{01d}\label{01d}\\
\gamma_\upsilon & \boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{01e}\label{01e}
\end{align}
Then using the Figure he proved that $\phi\left(\boldsymbol{\upsilon}\right)\boldsymbol{=+}1$ based on the assumption that $\,\rm S_1\,$ is moving with respect to $\,\rm S_2\,$ with velocity $\,\boldsymbol{-\upsilon}$. If this assumption is false then the Lorentz transformation is false, the in-variance of the speed $\,c\,$ of light in empty space in all inertial frames is not valid, so the co-variance of the Maxwell equations is not valid.
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
$\boldsymbol{\S}$ B. From the validity of the Lorentz Transformation
If we agree on the validity of the Lorentz Transformation between the inertial frames $\rm S_1,\rm S_2$ of the Figure, where $\rm S_2$ is moving along the common $x\boldsymbol{-}$axis with algebraic magnitude of the velocity $\upsilon \in [0,c)$ then
\begin{equation}
\mathbf R_2 \boldsymbol{=}
\begin{bmatrix}
x_2 \vphantom{\dfrac{a}{b}}\\
y_2 \vphantom{\dfrac{a}{b}}\\
z_2 \vphantom{\dfrac{a}{b}}\\
ct_2 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\gamma_\upsilon & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \boldsymbol{-} \gamma_\upsilon \dfrac{\upsilon}{c}\vphantom{\dfrac{a}{b}}\\
0 & \hphantom{\gamma_\upsilon} 1 & \hphantom{\gamma_\upsilon} 0 & 0\vphantom{\dfrac{a}{b}}\\
0 & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 1 & 0\vphantom{\dfrac{a}{b}}\\
\boldsymbol{-} \gamma_\upsilon \dfrac{\upsilon}{c} & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \gamma_\upsilon\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\begin{bmatrix}
x_1 \vphantom{\dfrac{a}{b}}\\
y_1 \vphantom{\dfrac{a}{b}}\\
z_1 \vphantom{\dfrac{a}{b}}\\
ct_1 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}
\mathrm L_{12}\mathbf R_1
\tag{02}\label{02}
\end{equation}
Now, suppose that $\rm S_1$ is moving with respect to $\rm S_2$ along the common $x\boldsymbol{-}$axis with algebraic magnitude of the velocity not $\upsilon^\prime\boldsymbol{=-}\upsilon$ but of $\upsilon^\prime\boldsymbol{=-}\left(\upsilon\boldsymbol{+}\Delta \upsilon\right)$. Then
\begin{equation}
\mathbf R_1 \boldsymbol{=}
\begin{bmatrix}
x_1 \vphantom{\dfrac{\upsilon^\prime}{c}}\\
y_1 \vphantom{\dfrac{\upsilon^\prime}{c}}\\
z_1 \vphantom{\dfrac{\upsilon^\prime}{c}}\\
ct_1 \vphantom{\dfrac{\upsilon^\prime}{c}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\gamma_{\upsilon^\prime} & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \boldsymbol{-} \gamma_{\upsilon^\prime}\dfrac{\upsilon^\prime}{c}\vphantom{\dfrac{a}{b}}\\
0 & \hphantom{\gamma_\upsilon} 1 & \hphantom{\gamma_\upsilon} 0 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\
0 & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 1 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\
\boldsymbol{-} \gamma_{\upsilon^\prime}\dfrac{\upsilon^\prime}{c} & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \gamma_{\upsilon^\prime}\vphantom{\dfrac{\upsilon^\prime}{c}}
\end{bmatrix}
\begin{bmatrix}
x_2 \vphantom{\dfrac{\upsilon^\prime}{c}}\\
y_2 \vphantom{\dfrac{\upsilon^\prime}{c}}\\
z_2 \vphantom{\dfrac{\upsilon^\prime}{c}}\\
ct_2 \vphantom{\dfrac{\upsilon^\prime}{c}}
\end{bmatrix}
\boldsymbol{=}
\mathrm L_{21}\mathbf R_2
\tag{03}\label{03}
\end{equation}
From equations \eqref{02} and \eqref{03}
\begin{equation}
\mathbf R_2 \boldsymbol{=}
\left(\mathrm L_{12}\mathrm L_{21}\right)\mathbf R_2
\tag{04}\label{04}
\end{equation}
This means that we must have
\begin{equation}
\mathrm L_{12}\mathrm L_{21} \boldsymbol{=}\mathrm I_{4}\boldsymbol{\equiv}\text{the identity $4\times 4$ matrix}
\tag{05}\label{05}
\end{equation}
that is
\begin{equation}
\!\!\!\!\!\!\!\!\begin{bmatrix}
\gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right) & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \boldsymbol{-} \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(\dfrac{\upsilon }{c} \boldsymbol{+}\dfrac{\upsilon^\prime}{c}\right)\vphantom{\dfrac{a}{b}}\\
0 & \hphantom{\gamma_\upsilon} 1 & \hphantom{\gamma_\upsilon} 0 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\
0 & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 1 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\
\boldsymbol{-} \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(\dfrac{\upsilon }{c} \boldsymbol{+}\dfrac{\upsilon^\prime}{c}\right) & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)\vphantom{\dfrac{\upsilon^\prime}{c}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
1 & 0 & 0 & 0 \vphantom{\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)}\\
0 & 1 & 0 & 0 \vphantom{\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)}\\
0 & 0 & 1 & 0 \vphantom{\dfrac{a}{b}}\\
0 & 0 & 0 & 1 \vphantom{\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)}
\end{bmatrix}
\tag{06}\label{06}
\end{equation}
Equation \eqref{06} is valid if
\begin{equation}
\gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(\dfrac{\upsilon }{c} \boldsymbol{+}\dfrac{\upsilon^\prime}{c}\right)\boldsymbol{=}0 \quad \textbf{and} \quad \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)\boldsymbol{=}1
\tag{07}\label{07}
\end{equation}
The conditions \eqref{07} are satisfied if and only if
\begin{equation}
\boxed{\:\:\upsilon^\prime\boldsymbol{=-}\upsilon \vphantom{\dfrac{a}{b}}\:\:}
\tag{08}\label{08}
\end{equation}
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
(1)
This Figure is extracted from my answer here : Schutz's geometrical proof that spacetime interval is invariant.