We need to find a transformation that maps the segments. It is sufficient to restrict the search to affine maps and to map only the end points
$$
z_1=\left[\matrix{x_1\\y_1}\right]\ \mapsto w_1=\left[\matrix{x_3\\y_3}\right],\quad\text{and}\quad
z_2=\left[\matrix{x_2\\y_2}\right]\ \mapsto \ w_2=\left[\matrix{x_4\\y_4}\right].
$$
Case I: $\text{rank}\,[z_1\ z_2]=2$. Then the linear transformation
$$
w=Az\qquad\text{where}\qquad A=[w_1\ w_2]\cdot [z_1\ z_2]^{-1}
$$
does the job.
Case II: $z_1\parallel z_2$, but $z_1\ne z_2$. Then we need to consider an affine map $w=Az+z_0$ to shift the $w$-vectors. Let $\lambda z_1+\mu z_2=0$. Then
$$
\left\{
\begin{array}{lcl}
\lambda Az_1+\lambda z_0&=&\lambda w_1,\\
\mu Az_2+\mu z_0&=&\mu w_2,\\
\end{array}
\right.\quad\Rightarrow\quad \underbrace{A(\lambda z_1+\mu z_2)}_{=0}+(\lambda+\mu)z_0=\lambda w_1+\mu w_2.
$$
It gives $z_0=\frac{\lambda w_1+\mu w_2}{\lambda+\mu}$. Note that $\lambda+\mu\ne 0$ by the assumption that $z_1\ne z_2$. With this $z_0$ any $A$ that maps only one end point, i.e. $Az_1=w_1-z_0$ would suffice (not unique).
Case III: $z_1=z_2$ is trivial. If $w_1=w_2$ (point-to-point) then any $A$ with $Az_1=w_1$ works. If $w_1\ne w_2$ (point-to-segment) then no such transformation (even non-linear) exists.
I am not an expert and have just starting thinking about this myself. I am intrigued by how many different ways there are to think about transforms / degrees of freedom.
I think the simplest way to see that an Affine transform has 6 degrees of freedom is that there are 6 variables in the matrix:
$$
\begin{bmatrix}
m_{00} & m_{01} & m_{02} \\
m_{10} & m_{11} & m_{12} \\
0 & 0 & 1 \\
\end{bmatrix}
$$
No matter what value we choose for any of those variables, it is a valid Affine transform. Although the Similarity transform can also be represented by a 6 variable multiplication matrix, it is more constrained - if we picked 4 of the variables at random, the other 2 we would have to choose carefully in order that it is a valid Similarity transform. So it has less degrees of freedom even though it still can be written as a matrix with 6 variables. Similarly, we can use an Affine transform to describe a simple translation, as long as we set the four left numbers to be the identity matrix, and only change the two translation variables.
The purest mathematical idea of an Affine transform is these 6 numbers and the way you multiply them with a vector to get a new vector. What this transform actually does can be described in a variety of ways - as 6 operations that you are doing one after the other (translate x, translate y, scale x, scale y, rotate, shear), or one thing you are doing all at once. If you think of them in terms of these operations, you might be confused by this matrix:
$$
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
This matrix can be thought of as either a rotation by 180 degrees about the origin, or of scaling x by -1 and y by -1, or by reflecting x and y through the origin. All of the transformations are equivalent, and this is the only matrix that describes them.
Another way we could think about degrees of freedom is with how many fingers you would need to describe this transform by dragging points.
A translation I can describe with one finger - by dragging a single point to its new location. Open Google Maps on your phone and try it. Each finger counts for two degrees of freedom since you can move it horizontally, and vertically.
A euclidean transform has 3 DOF - you need one finger to translate the shape, then the second finger you can use to rotate it, but this finger only has one degree of freedom. This one is better illustrated not in Google maps, but with a credit card on a desk - one finger moves the card, the other rotates it, but the second finger is less free since it always has to follow the first finger around somewhat. Moving the second finger arbitrarily would try to stretch the card, which is impossible. So, the first finger has 2 DOF, the second finger has one more.
A similarity transform has four degrees of freedom - Google Maps works for this one again. Drag two fingers on your phone on Google Maps at the same time. No matter where you drag your two fingers, the app is able to find a similarity transform for you - one that keeps the map the same shape, but translates, rotates, and scales it.
You would need to drag three fingers to do an Affine transform - Google Maps doesn't support this, since it would skew the map, so you wouldn't be able to navigate it using it anymore - but you can kind of pretend using a hankerchief, two of the fingers can translate and rotate it (pretend they can scale it too) and then the third finger can skew it this way and that. Almost any drag
And, dragging four fingers would let you do a 2D homogenous translation. You can try this in a photo editing program called Gimp. It's under tools > transform tools > perspective, and it lets you drag four different points around - so it counts for 8 degrees of freedom.
Note that not every possible position of the 4 points is necessarily a valid transform, but it still counts as 8 degrees of freedom since the points can still freely move in 2 dimensions - there's just certain values they can't take.
Who knows, I hope this helps!
Best Answer
It’s fairly easy to construct an affine map that will send one line segment to the other, but two pairs of points aren’t enough to specify this map uniquely. Since you’re working in $\mathbb{RP}^3$, you’ll need another two pairs to nail down the mapping completely.
I’ll assume the common computer graphics convention of points being represented by row vectors and right-multiplication by a matrix to apply a transformation.
The construction relies on the fact that the columns of a transformation matrix are the images of the basis. So, start by forming the matrix $Q$ that has the homogeneous coordinates of the destination points $Q_k$ as rows. These are the images of the corresponding input points $P_k$, so perform a change of basis to the standard basis by left-multiplying by the inverse of the matrix with the $P_k$ as its rows. I.e., the transformation matrix is $$P^{-1}Q=\begin{bmatrix}P_1\\P_2\\P_3\\P_4\end{bmatrix}^{-1}\begin{bmatrix}Q_1\\Q_2\\Q_3\\Q_4\end{bmatrix}.$$ This requires the $P_k$ to be linearly independent as elements of $\mathbb R^4$, of course, so that they form a basis, which means that the four points can’t be coplanar in $\mathbb{RP}^3$. If you like, you can avoid the matrix inversion and multiplication by computing this product via row-reduction. Form the matrix $\left[\,P\mid Q\,\right]$ and row-reduce to produce $\left[\,I\mid P^{-1}Q\,\right]$.
You have $P_1(-1/2,0,0,1)$, $P_2(1/2,0,0,1)$ and their images $Q_1$ and $Q_2$, the endpoints of the destination line segment. Choose $P_3$ and $P_4$ so that $P$ is nonsingular. With fixed choices for the four source points you can then precompute $P^{-1}$. A convenient choice is $P_3(0,1,0,1)$ and $P_4(0,0,1,1)$, with which $$P^{-1}=\begin{bmatrix}-1&1&0&0\\-\frac12&-\frac12&1&0\\-\frac12&-\frac12&0&1\\\frac12&\frac12&0&0\end{bmatrix}.$$
The choice of $Q_3$ and $Q_4$ determines how the transformation behaves for points that aren’t on the source line. If you don’t care about that, some simple computational choices are to collapse the rest of the space by setting them both equal to zero, or by setting them equal to $P_3$ and $P_4$, respectively. The latter choice is well-behaved most of the time, but results in a singular transformation if either of these points lies on the destination line.
By the way, depending on what it is you’re doing, mapping one line segment to another directly could be a much easier way to go. The source line segment can be parameterized as $(1-t)P_1+tP_2=(t-\frac12,0,0,1)$. The corresponding point on the destination segment is then simply $(1-t)Q_1+tQ_2$. If you examine the transformation constructed above, you’ll see that this is exactly what it’s doing for points along the source line: $(t-\frac12,0,0,1)P^{-1}=(1-t,t,0,0)$, and multiplying $Q$ by this vector yields $(1-t)Q_1+tQ_2$. (In fact, this is true regardless of what you choose for $P_3$ and $P_4$—the first and last rows of $P^{-1}$ turn out to be independent of this choice.)