There are many approaches to deriving the Lorentz transformation. The two main ones are, I think:
Method 1. Assume the Minkowski metric $\eta = {\rm diag}(-1,1,1,1)$ and then define Lorentz transformations $\Lambda$ as that set of transformations which satisfy $\Lambda ^T \eta \Lambda = \eta$ (where I am using matrix notation).
Method 2. Quote postulates and reason from them.
My question concerns the latter method. There is a long history of arguments over exactly what has to be assumed. For the sake of my question I shall solve the question of linearity by asserting that I am interested in finding a linear transformation if there is one which satisfies whatever postulates I introduce. (Thus I am not interested in the question whether there may also be non-linear transformations which also satisfy the postulates).
I would like to suggest that in order to obtain the standard Lorentz transformation, it is sufficient to assume linearity and just two further things:
Postulate 1 (The Principle of Relativity). The motions of bodies included in a given space are the same among themselves, whether that space is at rest or moves uniformly forward in a straight line. (This implies that the mathematical form of laws of motion is unchanged from one inertial frame to another.)
Postulate 2. There is a finite maximum speed for signals (where a signal is an influence which can transmit a cause to an effect).
But I am aware that this is disputed because, it is asserted, one must also add a further assumption about the isotropy of space.
For clarity, I will first briefly present the undisputed part of the argument. One considers two frames in relative motion, with aligned axes, such that frame $S'$ proceeds at speed $v$ in the $x$ direction as observed in frame $S$. One first argues that coordinates $t'$ and $x'$ of any given event are functions of $t$ and $x$ alone, and since we have decided to seek a linear transformation, we may write $$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} \mycolv{ct'\\x'} = \mycolv{a & b\\d & e} \mycolv{ct\\x} $$ where $a,b,d,e$ are functions of $v$, to be determined. First one reasons that the very meaning of speed $v$ is that the events $(t',0)$ in S' correspond to $(t,vt)$ in S, from which one finds $d = - (v/c) e$. One also asserts, from Postulate 1, that speed just means relative speed so one must equally find that the events $(t,0)$ in S correspond to the events $(t',-vt')$ in S', from which it follows that $d = - (v/c) a$. Then one can invoke Postulate 2 to assert that the events $(ct,ct)$ in S must correspond to $(ct',ct')$ in S', which gives $a + b = d + e$.
Putting all the above together, one finds the transformation has to be of the form $$ \Lambda(v) = a(v) \mycolv{1 & \!-v/c\\-v/c & 1}, $$ where the function $a(v)$ is still to be discovered.
It is at this point that the question of isotropy comes in. It is entirely reasonable to propose that $\Lambda(v)^{-1} = \Lambda(-v)$. The question is whether this statement is forced upon us by the Principle of Relativity and the assumption of linearity. In a quite thorough treatment by Selene Routley here: [ https://physics.stackexchange.com/questions/253356/homogeneity-and-isotropy-and-derivation-of-the-lorentz-transformations ] it is asserted that $\Lambda(v)^{-1} = \Lambda(-v)$ is not forced upon us but amounts to a further assumption. My question is to ask: is that correct?
Another way to ask the same question is: is it possible to furnish a linear transformation with $\Lambda(v)^{-1} \ne \Lambda(-v)$ which nevertheless satisfies the Principle of Relativity?
To make the question more pointed still, here is an argument to derive $a(v) = (1-v^2/c^2)^{-1/2}$ (and hence $\Lambda(v)^{-1} = \Lambda(-v)$) from the two postulates above, together with linearity, and no further assumption. The argument is essentially the one proposed by udrv here [ https://physics.stackexchange.com/questions/230320/how-do-the-postulates-of-relativity-relate-lorentz-transforms-to-their-inverses ].
We consider frames S' and S as before, and a further frame S'' which moves at speed $u$ in the $x'$ direction relative to frame S'. We argue from relativity postulate that S'' is just another inertial frame so we must find there is a transformation directly from S to S'' which must match the composition of the two transformations from S to S' to S'': $$ a(w) \mycolv{1 & -w/c \\ -w/c & 1} = a(u) \mycolv{1 & -u/c \\ -u/c & 1} a(v) \mycolv{1 & -v/c \\ -v/c & 1}. \tag{1} $$ By multiplying out the matrix product one finds $$ a(w) = a(u) a(v) (1 + u v/c^2) \;\;\; {\rm and} \;\;\; w = \frac{u+v}{1+uv/c^2} $$ hence the function $a(v)$ must satisfy $$ a\left( \frac{u+v}{1+uv/c^2} \right) \equiv a(u) a(v) \left( 1 + \frac{uv}{c^2} \right). \tag{2} $$ This is not just an equation but an identity: that is, it is valid at all $u,v$ in the range over which those quantities are defined, i.e. here $0 \le |u| < c$ and $0 \le |v| < c$. Such an identity is sufficient to fix the function uniquely. (To prove this, one way is to write $a(x) = \sum_{i=0}^\infty b_i x^i$ with constant coefficients $b_i$ to be discovered, and then make the claim that equation (2) is sufficient to fix the $b_i$). One obtains $$ a(v) = \frac{1}{\sqrt{1 - v^2/c^2}} $$ and hence we get the Lorentz transformation, without requiring any assumption about isotropy of space.
It would be helpful if any answer which asserts that isotropy is a further assumption could also explain why equation (1) does not itself follow from the two postulates and linearity.
