What does an automorphism of $\mathbb{RP}^1$ preserve that a homeomorphism does not? If we define an automorphism of $\mathbb{RP}^2$ to be a bijective self-map that preserves collinearity -- that is, a bijection that takes lines to lines -- then this property, all by itself, ensures that every automorphism of $\mathbb{RP}^2$ is represented by a $3 \times 3$ invertible matrix; this matrix is unique up to a scalar, and hence we can conclude that $Aut(\mathbb{RP}^2) \cong PGL(3,\mathbb R)$.
If we drop one dimension lower, though, the requirement that an automorphism must take lines to lines becomes trivial: $\mathbb{RP}^1$ is a single (projective) line, so any homeomorphism (indeed, any permutation) $\sigma: \mathbb{RP}^1 \to \mathbb{RP}^1$ preserves collinearity.  But not every bijective self-map $\mathbb{RP}^1 \to \mathbb{RP}^1$ is an automorphism.
What property of $\mathbb{RP}^1$ is preserved by an automorphism but not, in general, preserved by a homeomorphism of the underlying topological space?
To paraphrase the question, I would like to be able to fill in the blanks in the following statement:

An automorphism of $\mathbb{RP}^1$ is a homeomorphic self-map such that _____________________.  We can prove that any such map is represented by a $2 \times 2$ invertible matrix, unique up to a scalar, and hence we conclude that $Aut(\mathbb{RP}^1) \cong PGL(2, \mathbb R)$.

Of course one way to fill in the blank is by saying "An automorphism of $\mathbb{RP}^1$ is a self-homeomorphism that can be represented by a fractional linear transformation of the form $t \mapsto \frac{at+b}{ct+d}$", but that is begging the question.  Why should we only consider maps to be automorphisms if they have this specific form?  For example, why is the map given by (in homogeneous coordinates) $(x, y) \mapsto (x^3, y^3)$ not an automorphism? What property does it lack?
I would prefer not to receive answers that begin with a general definition of an automorphism for projective $n$-space over an arbitrary field $k$, and then show that in the special case $n=1, k = \mathbb R$ it follows that $Aut(\mathbb{RP}^1) \cong PGL(2, \mathbb R)$. I am seeking an elementary perspective, one that need not generalize, that can help explain exactly what we are talking about when we talk about automorphisms of the real projective line.
 A: The standard answer to this is that automorphisms of $\mathbb{P}^1$ (over any field) are the bijections that preserve the cross-ratio.  The cross-ratio is the operation on 4-tuples of distinct points defined by $$(a,b;c,d)=\frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ (This definition obviously makes sense when $a,b,c,d$ are all finite; if one of them is infinite, you just say the two factors involving $\infty$ formally cancel out.  Or, in homogeneous coordinates, if you write $a=[a_0,a_1]$ and similarly for the other variables, the cross-ratio can be written as $[(a_0c_1-a_1c_0)(b_0d_1-b_1d_0),(a_0d_1-a_1d_0)(b_0c_1-b_1c_0)]$.)
One slick way to see this is to observe that for any distinct $b,c,d\in\mathbb{P}^1$, there is a unique $f\in PGL(2)$ such that $f(b)=1$, $f(c)=0$, and $f(d)=\infty$, namely $f(t)=k\frac{t-c}{t-d}$ where $k=\frac{b-d}{b-c}$ is the constant that makes $f(b)=1$.  But now notice that this formula for $f(t)$ is exactly the cross-ratio $(t,b;c,d)$.  That is, $(t,b;c,d)$ is exactly the image of $t$ under the unique automorphism of $\mathbb{P}^1$ that sends $b$ to $1$, $c$ to $0$, and $d$ to $\infty$.  (If one of $b,c,$ or $d$ is $\infty$, then this calculation requires some minor modification, but you reach the same conclusion.)  It follows immediately that every automorphism of $\mathbb{P}^1$ preserves the cross-ratio.  Conversely, suppose $f:\mathbb{P}^1\to\mathbb{P}^1$ is a bijection that preserves the cross-ratio.  Let $b=f^{-1}(1)$, $c=f^{-1}(0)$, and $d=f^{-1}(\infty)$.  Then for any $t\neq0,1,\infty$, we must have $(f(t),1;0,\infty)=(t,b;c,d)$.  But $(f(t),1;0,\infty)$ is just $f(t)$, so this says $f(t)=(t,b;c,d)$ which is the formula for an element of $PGL(2)$.
