It is relatively easy to get constant speed of light as constant of nature from Galilean equivalence of inertial observers ( GP) principle.
GP says that any two observers, moving wit (any) constant speed relatively to each other, are equivalent, that is, they both describes physical reality in the same way.
At particular it means they share the same kinematics principles, and they are using isomorphic coordinate frames. That's is if observer O has this coordinate system $(t,x)$ and observer O' has $(t',x')$, they are related. If we consider the simplest relationship, linear dependence, it is described by the following equations valid for observer O' :
$$x' = at + bx$$
$$t' = ct + dx$$
As observer O is equivalent to observer O' and physical reality observed from coordinate frame O is exactly the same, the same relations must be valid for observer O :
$$x = at' + bx' $$
$$t = ct' + dx' $$
Please, notice, the same coefficients a,b,c,d are used!
If O and O' has synchronized clocks, we may combine both descriptions, trying to obtain further requirements for coefficients a,b,c,d, as described ( in Polish), here. Assume that O and O' moves with relative speed V.
We have to define ( unknown) function d( V)
$$x = d( x' + Vt' )$$
$$t = d( t' + x' V \frac{1-d^{-2}}{V^2})$$
Now, if we assume there is third observer O'' which moves with speed of U relatively to observer O' we may ask, what is her speed for observer O? Using above formulas we obtain the following result:
$$ O("U+V") = \frac{U+V}{1+UV\frac{1+d^{-2}(V)}{V^2}} $$
If we set that $d(V) =1 $, we will have Galilean physics restored. But there is no argument to do this!
We where using observer O as a basis, and observer O' moving with speed V relatively to it. Another observer O'' was moving with the speed of U relatively to O', and we obtain related speed of O'' valid for observer O. Speed U was speed of O'' in frame O' which was "dragged" by frame O' wirth "draging speed" of V.
What if we would start with observer O'' and calculate the same? As O and O' are equivalent, the only difference is, that this time, "dragging speed" will be U whilst another one, V would be relatively to O'. In other words, U and V exchange its role. It means, that relative speed $O("U+V")$ and $O''("V+U")$ Has to be the same! Observers are equivalent, remember!
So we may write:
$$ \frac{U+V}{1+UV \frac{1+d^{-2}(V)}{V^2}} = \frac{V+U}{1+VU\frac{1+d^{-2}(U)}{U^2}} $$
After rearranging terms we obtain the following formula:
$$ \frac{1+d^{-2}(V)}{V^2} = \frac{1+d^{-2}(U)}{U^2} $$
There is only one way to fulfill the following equation: $f(x)=f(y)$. $f(x)$ has to be constant! So the whole fraction.
We obtain the final result:
$$\frac{1+d^{-2}(V)}{V^2} = C$$
where from dimensional analysis cames that constant C has dimension of 1/ velocity^2. It has to be universal constant assuming validity of GP, and strict Galilean physics is restored when C is equal to 0.
Function $d(V)$ is as follows:
$$d(V)=\frac{1}{\sqrt{(1-CV^2)}}$$
In this way we obtain Lorentz transformations assuming GP and following general rules with linear relationship between three inertial observers.
If we want to know value of ( unknown yet) constant C we should perform various experiments. But we may notice, that Lorenz transformations leaves Maxwell equations unchanged, and it gives us relation between our constant C and speed of light in vacuum:
$$C=\frac{1}{c^2}$$
Reasoning above was performed ( and published in 60'ties) by polish physicist Andrzej Szymacha.