Is the Lyness 5-cycle map birationally conjugate to its own square? Let $L(x,y) = (y,(y+1)/x)$. On a dense open subset of the plane, $L$ and all its powers are well-defined invertible maps and $L^5$ is the identity ($L$ is sometimes called the Lyness 5-cycle map). $L$ is birationally conjugate to its own inverse via the map that exchanges $x$ and $y$. Is $L$ furthermore birationally conjugate to $L^2$? That is, is there some birational map $f$ satisfying $f^{-1} \circ L \circ f = L^2$?
My past MO post Conjugating the Lyness 5-cycle into a rotation of the plane is relevant, and Brunault's reply would give me affirmative answer to my question if I knew that the rotation map of the projective plane associated with $L$ is birationally conjugate to its own square. (I briefly thought Galois theory would bridge this gap, but it smells too much like the problem of extending automorphisms of finite extensions of ${\bf Q}$ to automorphisms of ${\bf C}/{\bf Q}$.)
In their article "On Cremona transformations of prime order" (https://arxiv.org/pdf/math/0402037.pdf), Beauville and Blanc (summarizing the state of knowledge prior to their work) write "The linear transformations of a given order are contained in a unique conjugacy class." And apart from these, among other conjugacy classes: " ... As for conjugacy classes of order 5, there is at least one family, given by automorphisms of a special Del Pezzo surface of degree 1." Then they state their main result (Theorem 1): In the group of birational automorphisms of the projective plane, every conjugacy class of order 5 is either "of the above type" or is the class containing the linear automorphisms.
It is not clear to me whether this is asserting that there are exactly two conjugacy classes of order-5 automorphisms (one containing all the Del Pezzo automorphisms of order 5 and one containing all the linear automorphisms of order 5).
 A: The Del Pezzo connection gives another way to see that the map must be conjugate to its square and to construct an  explicit conjugation.  Blowing the plane up at four points in general position (i.e. no three collinear) yields the Del Pezzo quintic DP5 in projective 5-space, and DP5 has a group $S_5$ of automorphisms that contain the Lyness map as a 5-cycle.  Now observe that a 5-cycle is conjugate to its square in $S_5$, and thus a fortiori conjugate in the group of all birational automorphisms of the plane.
P.S. An explicit $f$ such that $L \circ f = f \circ L^2$ is
$$
f(x,y) = \left( -\,\frac{x}{x+1}, \, -\,\frac{y+1}{x+y+1} \right).
$$
Both $L \circ f$ and $f \circ L^2$ take $(x,y)$ to $-(y+1,x+1)/(x+y+1)$.
The map $f$ is birational because $f^4$ is the identity.
A: In their article On Cremona transformations of prime order, Beauville--Blanc complete the proof of the fact that there are exactly two conjugacy classes of birational automorphisms of $\mathbf{P}^2$ of order $5$: one coming from a special Del Pezzo surface of degree 1, and one containing all the linear automorphisms of $\mathbf{P}^2$ of order $5$. In particular, they prove that the automorphism of order $5$ of the Del Pezzo surface of degree 5 is conjugate to a linear automorphism of $\mathbf{P}^2$, so lies in the second conjugacy class.
Regarding your question, $L$ is conjugate to a linear automorphism $M$ of order $5$ of $\mathbf{P}^2$, so $L^2$ is conjugate to $M^2$, which also has order $5$, and Proposition on p.2 in Beauville--Blanc shows that $M$ and $M^2$ are conjugate in the Cremona group. It follows that $L$ and $L^2$ are conjugate in the Cremona group, and you should be able to compute the conjugating automorphism using Beauville--Blanc's last remark.
