How exactly are Lorentz transformations rotations? I've always seen everywhere that Lorentz transformations are rotations in 4D. Let's stick with 2D (one space axis, one time axis) for simplicity.
Rotations of two dimensional space axes look completely different from the 2D Lorentz transformation. To rotate space axes, we rotate both x and y axes by an angle of same magnitude with same sign. This results in the axes still remaining at 90 degrees after the rotation.
But Lorentz transformations in 2D look like rotating both space and time axes by an angle of same magnitude but opposite signs. The axes don't remain at 90 degrees after Lorentz transformations, instead they form a V-shape. How is this a rotation? Are we using some generalised definition of rotation? Also, why not just rotate both space and time axes in the same direction (so that they remain at 90 degrees), like we do with two space axes? (I've read the reason we don't treat the time-axis like a usual space-axis is that we can't move backwards in time. If this reason is correct, then please elaborate on this. How does rotating both x and t axes in the same direction mean we are going backwards in time?)
EDIT- https://www.mathpages.com/rr/s1-07/1-07.htm I'm coming off this text. Near the end of the first paragraph, it says that the reason we choose to do the 'opposite sign' rotation is that we can't go backwards in time.
 A: Wikipedia has some nice animations showing that Lorentz transformations are indeed some sort of rotation, e. g. this one

or this

One might say that they do their best to look like rotation under the constraint of not crossing the lightcone
A: The relevant generalized notion of "rotation" is that a rotation is a transformation that fixes one point and preserves all distances.  In euclidean space, this means that if you have two points with x-coordinates differing by $\Delta x$ and y-coordinates ďiffering by $\Delta y$, then the value of $\Delta x^2+\Delta y^2$ is unaffected by a rotation.  In Minkowski space it means that $\Delta x^2-\Delta t^2$ is unaffected.
A: A Lorentz boost is not a rotation by a real angle. Instead. it is a strain by a real angle.  The transformation of the x,t axis where both move inward by a small angle in radians $d\lambda$ (called the Lorentz boost parameter) is well known to mechanical engineers in the x,y plane.  The engineer distorts a square in the x,y plane so that both edges of the square move inward by a small angle in radians $d\epsilon$ (called the strain).  The square becomes a parallelepiped.  The matrices which do these transformations for non-infinitesimal angles are:
$$
\begin{bmatrix}
cosh(\lambda) & sinh(\lambda)\\
sinh(\lambda) & cosh(\lambda)\\ 
\end{bmatrix}
\begin{bmatrix}
x\\
ct\\ 
\end{bmatrix}
\quad 
and
\quad
\begin{bmatrix}
cosh(\epsilon) & sinh(\epsilon)\\
sinh(\epsilon) & cosh(\epsilon)\\ 
\end{bmatrix}
\begin{bmatrix}
x\\
y\\ 
\end{bmatrix}
$$
The reason you have heard that boosts are somehow rotations is that old time physicists made boosts look like familiar rotations by using imaginary angles and making t imaginary.
$$
\begin{bmatrix}
x'\\
ict'\\ 
\end{bmatrix}
=
\begin{bmatrix}
cos(i\lambda) & -sin(i\lambda)\\
sin(i\lambda) & cos(i\lambda)\\ 
\end{bmatrix}
\begin{bmatrix}
x\\
ict\\ 
\end{bmatrix}
$$
$$
\begin{bmatrix}
x'\\
ict'\\ 
\end{bmatrix}
=
\begin{bmatrix}
cosh(\lambda) & -i\ sinh(\lambda)\\
i\ sinh(\lambda) & cosh(\lambda)\\ 
\end{bmatrix}
\begin{bmatrix}
x\\
ict\\ 
\end{bmatrix}
$$
$$
\begin{bmatrix}
x'\\
ct'\\ 
\end{bmatrix}
=
\begin{bmatrix}
cosh(\lambda) & sinh(\lambda)\\
sinh(\lambda) & cosh(\lambda)\\ 
\end{bmatrix}
\begin{bmatrix}
x\\
ct\\ 
\end{bmatrix}
$$
Space-space parallelepiped strains leave $x^2-y^2$ invariant.  Space-time parallelepiped strains leave $x^2-(ct)^2$ invariant.  Rotations leave $x^2+(ict)^2$ invariant.  Please see my answer to this question if you would like more math.  
I don't think "the reason we don't treat the time-axis like a usual space-axis is that we can't move backwards in time" is a good argument. However, if $ct>x$ a boost can not make $ct'<x'$ because $x^2-(ct)^2$ is invariant.  Thus if an event is causal in the forward light cone then it is causal in the forward light cone in all boosted frames too. A real rotation of x and real ct could turn $ct>x$ into $ct'<x'$ and screw up causality.
