Why is this non-linear transformation not a Lorentz transformations? It does preserve $x^2 + y^2 + z^2 - c^2t^2 = x'^2 + y'^2 + z'^2 - c^2t'^2 $ The Lorentz transformation is a mapping of the $x', y',
z',$ and $t'$ coordinates relative to $x, y, z, t$ that preserves the speed of
light:
(1) $x^2 + y^2 + z^2 - c^2t^2 = x'^2 + y'^2 + z'^2 - c^2t'^2 = 0$
If $K'$ is moving with speed $v$ in the $x$ direction relative to $K$ (so that $K$ is moving with speed $-v$ in the $x$ direction relative to $K'$), and we use the Galilean transformation, we get an erroneous
(2) $x'^2 + y'^2 + z'^2 - c^2t'^2 = -2vxt + v^2t^2$
which is NOT equal to zero, and hence does not describe a sphere of
radius $ct$ emanating from the origin of $K'$, but something else (I'm
pretty sure it's a sphere that's been shifted negatively in the $x$
direction by a small amount, relative to $K'$). The Lorentz
transformation says we modify both the $x' = x - vt$ and the $t' = t$
relationships in order to ensure that we end up getting 0.
However, as a naive mathematician, I look at equation (2) and think,
"Ok, so we've got to modify the rule for $x'$, or $t'$, or both, to get
rid of that nasty $-2xvt + v^2t^2$ term in the $x'$ rule. Why not just
set $t'$ up so that instead of coming out to $-c^2t^2$, it actually comes
out to $-c^2t^2 + 2vxt - v^2t^2$, so that the extra terms cancel out?" I
naively went ahead with this idea and I got the following coordinate
transformation:
$$
x' = x - vt\\
y' = y\\
z' = z\\
t' = \frac{1}{c} \sqrt{(v^2 + c^2)t^2 - 2vxt}
$$
If you plug this into (1), you get 0 like you're supposed to.
Now, I haven't been able to find an explanation as to why this transformation is unsatisfactory. Of course, it's no longer linear - but is there some basic reason why the transformation must be linear? Obviously this transformation satisfies Einstein's postulate about the constancy of the speed of light for all inertial observers.
Presumably, then, it must violate the first postulate, regarding the equivalence of all physical phenomena for inertial observers - otherwise we'd need more information in order to derive the Lorentz transformation, which I think is supposed to be derivable from the two postulates alone.
So, in the end, I have one question I'm trying to answer:


*

*Does this transformation violate equivalence for inertial observers? Why? (ideally, someone could give me an example of a force that suddenly appears in my coordinate system that didn't already exist in the other coordinate system)

 A: 1.  Response to your specific example:
  Suppose you and I are both located at a point we choose to call the origin.  You are traveling at velocity $3/5$ with respect to me.  (I take $c=1$).
Our friend, one light year away according to me, flashes a light at time $t=1/2$. 
According to you, that light flashes at time $i\sqrt{13/50}$.  
So I see the light flashing at a real time, while you see it flashing at an imaginary time, whatever that means.  Presumably it means you don't see it flashing at all.  That's one way you know you're moving.    
2.  Response to the more general question of "why linear?": More fundamentally:  If we both watch a third person moving away from the origin at some other velocity, and if no forces are acting on that person, and if he passes through both $(x_1,t_1)$ and $(x_2,t_2)$ (according to me) then we must have $x_1/t_1=x_2/t_2$.  And if you must agree that he has no forces acting on him, then we must also have $x_1'/t_1'=x_2'/t_2'$.   These considerations will force the implication $$(x_1 t_2=x_2 t_1) \Rightarrow (x_1' t_2'=x_2' t_1')$$
which very much restricts the allowable transformations.
A: This is where it is good to distinguish (x,y,z,t) as the coordinates on Minkowski space vs in a tangent space at any point. The Minkowski space comes to you as a manifold, but the tangent space comes to you as a vector space (so it makes sense to ask for linearity). It is then a happy case that you can identify them. This is what the $\Delta x$ of @knzhou answer is trying to say by moving you to the vector space instead of the manifold. If you look at what categories your objects are in, what you can do to them becomes no choice at all.
A: Flat spacetime in special relativity is the same everywhere, so we must have translational symmetry in time and space. If we start with two events separated by $\Delta x^\mu$ in one frame, the separation $\Delta x'^\mu$ in a boosted frame should only depend on $\Delta x^\mu$, and not on $x^\mu$ itself. This is only satisfied for linear transformations. 
Using your transformations, observers would be able to identify a privileged point. This violates the principle of relativity even for observers at rest relative to each other, since it depends on where they put their origins.
A: There is a theorem due to Zeeman that under very mild assumptions proves that admissible transformations in special relativity are elements of the Lorentz group. This is treated e.g. in Naber's "The Geometry of Minkowski Spacetime".
To fix definitions, an observer is admissible if it has a 3-dimensional, right-handed, Cartesian spatial coordinate system based on an agreed unit of length and relative to which photons propagate rectilinearly at a fixed speed in any direction, as well as an ideal standard clock based on an agreed unit of time with which to provide a quantitative temporal order to the events on his worldline, in which light speed will be 1.
The mapping from the coordinates of one admissible observer to that of another one constitutes an invertible map
$$F:\mathbb R^4\rightarrow\mathbb R^4$$
The following assumption will be made: any two admissible observers agree on the temporal order of any two events on the worldline of a photon, i.e. if $x$ and $y$ are events in the coordinates of one observer, and $\hat x$ and $\hat y$ those for another one, then $x^0 - y^0$ (time coordinates) and $\hat x^0 - \hat y^0$ have the same sign (it is not demanded that the times are equal).
Let $v$ be the velocity vector of the photon (which is constant). Since in our units $c = 1$, we have $\|v\| = 1$. We have for any two points $x$ and $y$ on the world line of the photon that
$$x^i - y^i = v^i(x^0 - y^0)$$
for $i = 1,2,3$. Consequently, on the world line of the photon
$$(x^0 - y^0)^2 - (x^1 - y^1)^2 - (x^2 - y^2)^2 - (x^3 - y^3)^2 = 0$$
This is a cone in $\mathbb R^4$ with vertex at $y$. Obviously this cone must be preserved in any coordinate system, hence so does the change of coordinates map $F$. Zeeman called such a map a causal automorphism, and proved that $F = T\circ K\circ\Lambda$, where $T$ is a translation and $K$ is a dilatation of $\mathbb R^4$, and $\Lambda\in O(1,3)$ with $\Lambda^0_0\ge 1$.
Note that for this result it is not even assumed that $F$ is continuous.
