How do the postulates of relativity relate Lorentz transforms to their inverses? This question is about one of the assumptions involved when one is deriving the Lorentz transformation, for example under "Principle of Relativity" in Wikipedia's page on the derivations. Let me elaborate:
Consider two inertial frames $S$ and $S'$ in the stantard configuration, i.e., $S'$ moves with velocity $v$ along the x-axis of $S$. Let us forget about the directions $y$ and $z$. The coordinates $(x,t)$ of $S$ and $(x',t')$ of $S'$ are related by (due to the Inertia law and first postulate of special relativity):
$$\begin{align}
x' &= \alpha_1 x+\alpha_2 t \tag{1} \\
t' &= \alpha_3 x+\alpha_4 t\tag{2}
\end{align}$$
and we want to fix $\alpha_i$.
The first step is to consider that a particle at rest in $S'$ is moving with velocity $v$ in $S$. Then,
$$x=vt\quad \text{if}\quad x'=0$$
which fixes 
$$\alpha_2=-\alpha_1 v\tag{3}$$
We use now the second postulate, i.e.,
$$\frac{x}{t}=\frac{x'}{t'}=c$$
which, together with (1) and (2), allows us to fix more one $\alpha$, let us say $\alpha_4$:
$$\alpha_4=\left(1-\frac{v}{c}\right)\alpha_1-c\alpha_3\tag{4}$$
We now claim the first postulate: we can consider that the reference frame $S$ is moving with velocity $-v$ with respect to $S'$. Now, the coordinates would be related by,
$$\begin{align}
x &= \tilde\alpha_1 x'+\tilde\alpha_2 t' \tag{5} \\
t &= \tilde\alpha_3 x'+\tilde\alpha_4 t' \tag{6}
\end{align}$$
Now, we have the condition
$$x=0\quad \text{if} \quad x'=-vt'$$
which fixes 
$$\tilde\alpha_2=\tilde\alpha_1 v\tag{7}$$
Using again the second postulate, we obtain now,
$$\tilde\alpha_4=\left(1+\frac{v}{c}\right)\tilde\alpha_1-c\tilde\alpha_3\tag{8}$$
Finally, we assume that the transformations (1),(2) and (5),(6) are inverses of each other. This would fix all the missing $\alpha_i$ and $\tilde\alpha_i$, except for one of them. This means, we can write all  $\alpha_i$ and $\tilde\alpha_i$ in terms of, let us say, $\alpha_1$.
Now comes the part I cannot understand: one says that by the first postulate we actually have:
$$x=\alpha_1 (x'+ v t')\tag{9}$$
i.e., we have $\tilde\alpha_1=\alpha_1$, which would fix then all the constants and we would obtain the Lorentz transformation.
However, for me it looks like that the first postulate only leads to (5), not to (9). Can someone elaborate on this? One argument seems to be just change "prime" to "not-prime", but this is not convincing me; I can not see why the first postulate implies $\tilde\alpha_1=\alpha_1$.
Another way to look at the problem
The coefficients $\alpha_i$ can be fixed by means of the following assumptions (no need to use the transformation related to  $\tilde\alpha_i$):
Assumption 1: The following holds:
$$
x^2-c^2t^2=x'^2-c^2t'^2
$$
Assumption 2: the coefficients $\alpha_i$ do not depend on $x$ and $t$.
Then, the question can be reformulated as: how Assumption 1 and  Assumption 2 follows from the Postulates of special relativity?
An old book
I have found an old reference: The Theory Of Relativity (1914) by L. Silberstein (the book can be read online). In this book, the author seems to claim that indeed $\alpha_1=\tilde\alpha_1$ is kind of arbitrary, see the discussion culminating on page 110. I would need to read it more carefully to confirm that this is indeed what he is saying, but I am mentioning the reference in case someone already did it.
 A: No need to force $\tilde\alpha_1 = \alpha_1$.
After taking advantage of the fact that (1), (2) and (5), (6) must be inverses of each other, your transformation reads, in matrix form,
$$
\left(\begin{array}{c} x'\\ t' \end{array}\right) = \alpha_1(v)\left(\begin{array}{cc} 1 & -v\\-\frac{v}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right)
$$
where the dependence of $\alpha_1$ on the relative velocity $v$ is shown explicitly, and if we identify $\tilde\alpha_1 = \alpha_1(-v)$ we also have
$$
\alpha_1(v) \alpha_1(-v) \left( 1 - \frac{v^2}{c^2}\right) = 1
$$
Consider now 3 distinct inertial frames, $S$, $S'$, and $S"$. Let $S"$ move at velocity $w$ relative to $S$, and denote $W(w)$ the corresponding coordinate transformation. Similarly, let the relative velocities of $S'$ and $S$, and of $S'$ and $S"$ be $u$ and $v$ respectively, and denote by $V(v)$ and $U(u)$ the corresponding coordinate transformations from $S$ to $S'$ and from $S'$ to $S"$. By the principle of relativity transformations $U(u)$, $V(v)$, and $W(w)$ must have the same form and the same dependence on the respective relative velocities. In addition, by the same principle of relativity, transforming coordinates from $S$ to $S"$ by means of $W(w)$ must produce the same result as transforming coordinates first from $S$ to $S'$ by means of $V(v)$ and then from $S'$ to $S"$ by means of $U(u)$. Otherwise we could have 2 different transformations from $S$ to $S"$, generating different results for at least some events, and we could easily use such a feature to distinguish between frames. It follows that necessarily 
$$
W(w) = U(u) V(v)
$$
The right hand side compound transformation amounts to
$$
\left(\begin{array}{c} x"\\ t" \end{array}\right) = \alpha_1(u)\alpha_1(v)\left(\begin{array}{cc} 1 & -u\\-\frac{u}{c^2} & 1 \end{array}\right)\left(\begin{array}{cc} 1 & -v\\-\frac{v}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right) 
$$
and doing the algebra gives eventually
$$
\left(\begin{array}{c} x"\\ t" \end{array}\right) = \alpha_1(u)\alpha_1(v)\left(1+\frac{uv}{c^2}\right)\left(\begin{array}{cc} 1 & -\bar w\\-\frac{\bar w}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right)
$$
where 
$$
\bar w = - \frac{u+v}{1 + \frac{uv}{c^2}}
$$
But this must be the same as 
$$
\left(\begin{array}{c} x"\\ t" \end{array}\right) = \alpha_1(w)\left(\begin{array}{cc} 1 & -w\\-\frac{w}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right)
$$
Equating the previous result element by element with this direct transformation shows that
$$
\alpha_1(w) = \alpha_1(u)\alpha_1(v)\left(1+\frac{uv}{c^2}\right)
$$
$$
- w \alpha_1(w) = - \bar w \alpha_1(u)\alpha_1(v)\left(1+\frac{uv}{c^2}\right)
$$
The first identity above gives a composition law for $\alpha_1$. Substituting this in the 2nd identity gives a form of the velocity addition rule, even though we don't yet know the explicit expression of $\alpha_1$:
$$
w = \bar w = - \frac{u+v}{1 + \frac{uv}{c^2}}
$$
Now notice that 
$$
w = v \;\;\Rightarrow \;\; u = - \frac{2v}{1+\frac{v^2}{c^2}}
$$
and substitute into the composition rule for $\alpha_1$:
$$
\alpha_1(v) = \alpha_1(u) \alpha_1(v) \frac{1 - \frac{v^2}{c^2}}{1+\frac{v^2}{c^2}}\;\;\Rightarrow\;\;\alpha_1(u) = \frac{1 + \frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}
$$
for $u = - \frac{2v}{1+\frac{v^2}{c^2}}$. But let us notice further that 
$$
1 \pm \frac{u}{c} = 1 \mp \frac{2v}{1+\frac{v^2}{c^2}} = \frac{\left(1 \mp \frac{v}{c} \right)^2}{1+\frac{v^2}{c^2}}
$$
Then we must also have
$$
1 - \frac{u^2}{c^2} = \left(1 - \frac{u}{c}\right)\left(1 + \frac{u}{c}\right) = \left(\frac{1- \frac{v^2}{c^2}}{1+\frac{v^2}{c^2}} \right)^2\;\;\Leftrightarrow\;\;
\frac{1 + \frac{v^2}{c^2}}{1-\frac{v^2}{c^2}} = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}
$$
and therefore
$$
\alpha_1(u) = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}
$$
Obviously $\alpha_1(-u) = \alpha_1(u)$ and the original condition $\alpha_1(u) \alpha_1(-u) \left( 1 - \frac{u^2}{c^2}\right) = 1$ is satisfied.
Note: The light speed condition isn't actually necessary. Homogeneity and isotropy of space also lead to Lorentz transformations through a couple of extra algebraic steps ($c$ enters as parameter and must be identified as light speed a posteriori), with a side alternative of Galilei transformations (corresponding to $c \rightarrow \infty$).
A: The first postulate of Special Relativity - which is also the Principle of Relativity is "Laws of Physics are invariant in all the inertial frames."
Now consider the following scenario:
Suppose there are two events which are happening simultaneously in frame $O$ and are spatially separated by distance $l$. From your Equation $1$, the spatial interval between the same events is $\alpha_1$$l$ in a frame $O'$ which is moving uniformly with respect to the frame $O$.
Now consider the reverse situation. There are another two events which are happening simultaneously in frame $O'$ and have spatial interval $l$ between them. Then from your Equation $5$, it is clear that the spatial interval between the same events is $\tilde\alpha_1l$ in frame $O$.
But from the principle of relativity, the laws of Physics must be same in all the inertial frames. So the factor by which the spatial interval between the simultaneous events of one frame gets multiplied when we see them from a different frame should be same for a given relative speed between the two frames. It is just the symmetry argument based on the symmetry already postulated in the principle of relativity. Thus, $\alpha_1$ must be equal to $\tilde\alpha_1$.
