Extending birational isomorphisms between planar curves to the P^2 Let $k$ be a field and let $C,D$ be two integral curves in $\mathbb{P}^2_k$. Now let $f:C \to D$ be a birational isomorphism. Can $f$ be extended to $\mathbb{P}^2_k$. To be precise, does there exist a birational isomorphism $F:\mathbb{P}^2_k \to \mathbb{P}^2_k$ such that $F$ and $f$ agree on a (non empty) open subset of $C$?
I am mainly interested in the case when $k=\mathbb Q$. 
[Edit] Since it is apparently easy to read over I will state it here explicitly. The map $F$ is allowed to be a birational isomorphism. It is clear that my statement is false when you want $F$ to be an isomorphism since it has to send curves of the same degree to curves of the same degree.
If we take for example $C$ to be the curve $x=0$ and $D$ to be the curve $y^2z=x^3+x^2z$ then $C$ and $D$ are birationally equivalent but clearly no automorphism of $\mathbb{P}^2_k$ sends C to D. However it is possible with a birational isomorphism.
 A: Another way to see that a nodal rational curve of degree $\ge 6$ cannot be sent to a line by a birational map of the plane is to see that it cannot be contracted by a birational map of the plane. Each birational map of the plane decomposes into blow-ups and contractions of smooth $(-1)$-curves (curves isomorphic to $\mathbb{P}^1$ and of self-intersection $-1$). If you want to contract the curve, you have to blow-up all nodes. The number of nodes is $(d-2)\cdot (d-1)/2$, and the self-intersection of the curve on $\mathbb{P}^2$ is $d^2$. After blowing-up, the self-intersection decreases by $4$ for each double point, so becomes $d^2-2\cdot (d-1)(d-2)=6d-d^2+4$. If $d\ge 6$, this number is $\le -2$, so the curve is not contractible. If $d\le 5$, the curve is contractible and in fact one can check easily that it can be sent onto a line by a birational map of the plane.
Similarly, one sees that a nodal rational curve of degree $d\ge 6$ cannot be sent onto a nodal rational curve of degree $d'\ge 6$ when $d\not=d'$.
Similar arguments work with curves of genus $1$. See for example  Proposition 3.3.3 of "On birational transformations of pairs in the complex plane", J. Blanc, I. Pan, T. Vust, Geom. Dedicata 139 (2009), 57-73.
A: EDIT: As  René Pannekoek points out in a comment, I misread the question.  This question was asking for birational automorphisms of $\mathbb{P}^2$, not just automorphisms.  
I'll leave this answer (which doesn't answer the question) since it might be useful.
A negative result for elliptic curves:  Not in general.  For example, suppose that $C$ and $D$ are the same elliptic curve inside $\mathbb{P}^2$.  Set $P \in C$ to be an inflection point, ie a point such that $L \cap C = 3P$ for some line $L \in \mathbb{P}^2$.  Fix $Q$ to be another point which is not of that form (ie, such that the tangent line intersects $Q$ at another point of $C = D$).  In particular, let's say that $Q$ is an element of infinite order.
Suppose that $f : C \to D$ to be a map which sends $P$ to $Q$ (note that such maps always exist since $C$ is an elliptic curve and thus an Abelian variety).  No automorphism of $\mathbb{P}^2$ (ie, an element of $PGL(2)$) will restrict to $f$, since such an $F$ will send the line $L$ going through $P$ to a line going through $Q$ (and only $Q$).
EDIT: As  François Brunault points out, this sort of construction can't work to avoid birational automorphisms.
A positive result for elliptic curves: On the other hand, if your elliptic curve is $C = V(y^2 - x(x-1)(x-\lambda) )$, with inflection point at infinity $P$, then Exercise 4.3 in Chapter IV of Hartshorne says that every isomorphism of $C$ that leaves $P$ fixed comes from a linear automorphism of $\mathbb{P}^2$.  
A positive result for genus 3 curves: The result is true for genus 3 curves, see Hartshorne, Chapter IV, Exercise 5.7(a).
A positive result vacuously: Of course, ``most'' curves of higher genus have no automorphisms at all (ie, most elements of the moduli space have no automorphisms except the identity).  See for example Baily, On the automorphism group of a generic curve of genus >2.
A: In general the answer is no.
This kind of question is studied in more generality in the paper by Mella and Polastri "Equivalent birational embeddings", Bull Lond. Math. Soc. 41 (2009) 89-93, http://arxiv.org/abs/0906.4858.
They prove that two birational embeddings of $X$ in $\mathbf{P}^n$ are equivalent up to Cremona transformations of $\mathbf{P}^n$ as long as $n\geq \dim(X)+2$. For instance, any rational variety of codimension at least $2$ in $\mathbb{P}^n$ is Cremona equivalent to a linear space.
The case $n=2$ and $\dim(X)=1$ is outside this range, and indeed there are examples of birational plane curves that are not equivalent up to Cremona transformations, hence not equivalent under the action of $\textrm{Bir}(\mathbb{P}^2)$, since $\textrm{Bir}(\mathbb{P}^2)$ is generated by Cremona transformations.
The following example is given by Mella and Polastri in the last section of their paper. Take a general projection of a curve of bidegree $(1,d)$ on a quadric surface to $\mathbb{P}^2$. This is a plane rational curve $C$ of degree $d$ with only ordinary double points, hence there is a birational isomorphism $C \dashrightarrow L$, where $L$ is a line. However, one proves that if $d \geq 6$ then $C$ is not Cremona equivalent to $L$. 
The proof is based on the following 

Lemma. Let $X \subset \mathbb{P}^n$ be a rational variety of codimension $1$ and degree $d>1$. If $X$ is Cremona equivalent to a hyperplane, then the singularities of the pair $(\mathbb{P}^n, \frac{n+1}{d} X)$ are not canonical. 

