Consider the Hom scheme $X=Hom(\mathbb P^1, \mathbb P^1)$ over an algebraically closed field $k$. Using deformation theory, I think we can show locally at a point representing a degree $n$ morphism, $X$ is smooth of dimension $2n+1$.
So if we consider the subscheme $X_n$ of $X$ consist of degree $n$ morphisms, it has a natural action by $X_1 \times X_1=Aut(\mathbb P^1) \times Aut(\mathbb P^1)= PGL(2) \times PGL(2)$ by composing morphism from left and right sides.
The question: is $X_n$ affine and connected? What is the quotient $X_n/PGL(2) \times PGL(2)$ ?
To give a map $\mathbb{P}^1 \to \mathbb{P}^1$ of degree $n$ is equivalent to giving a pair of two homogeneous polynomials of degree $n$ in two variables up to constant without common linear factors. This means that $X_n$ is an open subset of $$ \bar{X}_n = \mathbb{P}^{2n+1}. $$ This space $\bar{X}_n$ is called the space of quasimaps or the Drinfeld compactification of the space of maps.
Furthermore, it is clear that $\bar{X}_n \setminus X_n$ is the discriminant (of two polynomials) hypersurface, or equivalently the image of the natural map $$ \bar{X}_{n-1} \times \mathbb{P}^1 \to \bar{X}_n $$ that takes a pair of polynomials of degree $n-1$ and multiplies them by a common linear factor.
All this shows that $X_n$ is smooth and affine.
If $U$ and $V$ are two vector spaces of dimension 2 so that the source of the maps is $P(U)$ and the target is $P(V)$ then $$ \bar{X}_n = \mathbb{P}(V \otimes S^nU^*). $$ This isomorphism is $PGL(U) \times PGL(V)$-equivariant.