Does there exist a nonidentity (which also is not a rotation) rational map $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ with period $7$, i.e., for which the seventh iteration $f^7$ is the identity map?
I know such a function, $f:(x,y)\mapsto \left(\frac{1}{y},x(1+y)\right)$, with period $5$. Here is a verification done in Maple:
f := (x, y) ->( 1/y, x*(1+y));
(f@@5)(x, y);
$$\left( {\frac {1+x \left( 1+y \right) }{y} \left( 1+{\frac {1}{x \left( 1+y \right) } \left( 1+{\frac {1+x \left( 1+y \right) }{y}} \right) } \right) ^{-1}} \right. ,$$ $$\left.{x \left( 1+y \right) \left( 1+{\frac {y}{1+x \left( 1+y \right) } \left( 1+{\frac {1}{x \left( 1+y \right) } \left( 1+{\frac {1+x \left( 1+y \right) }{y}} \right) } \right) } \right) \left( 1+{ \frac {1+x \left( 1+y \right) }{y}} \right) ^{-1}}\right) $$
simplify((f@@5)(x, y)[1]),simplify((f@@5)(x, y)[2]);
$$ x,\,y$$
The $2 \times 2$ rotation matrix for an (anticlockwise) rotation by $\frac{\pi}{7}$ is $$\pmatrix{\cos \frac{\pi}{7} & -\sin \frac{\pi}{7} \\ \sin \frac{\pi}{7} & \cos \frac{\pi}{7}}.$$ Via the usual identification of that group with the group of Mobius transformations (and clearing the cosine factors for readability), the image of that matrix in $PSL(2, \Bbb R) = SL(2, \Bbb R) / \{\pm I\}$ under the canonical projection map can be identified with the rational function $$f_7 : t \mapsto \frac{t - \alpha}{1 + \alpha t},$$ where $$\alpha := \tan \frac{\pi}{7}.$$ Clearly $f_7 \neq \text{id}$ and via the above identification (or by repeated application of the tangent sum identity) we see that $f_7^7 = \text{id}$.
So, the rational transformation $$g_7 : (x, y) \mapsto (f_7(x), y)$$ is not the identity but satisfies $g_7^7 = \text{id}$.
We can produce a natural example that involves $y$ in a more essential way by replacing $t$ in the definition of $f_7$ with a complex variable $z = x + iy$ and decomposing $f_7$ into real and imaginary parts. This yields the rational map $$h_7 : (x, y) \mapsto \left( \frac{\alpha x^2 + (1 - \alpha^2) x - \alpha + \alpha y^2}{(1 + \alpha x)^2 + \alpha^2 y^2}, \frac{(1 + \alpha^2) y}{(1 + \alpha x)^2 + \alpha^2 y^2} \right)$$ of period $7$, which can be rewritten in various ways.
Note that this construction is not specific to the number $7$. Replacing it, for example, with $3$ and using the special value $\tan \frac{\pi}{3} = \sqrt{3}$ gives the analogous maps $$f_3 : x \mapsto \frac{x - \sqrt{3}}{1 + \sqrt{3} x}$$ and $$h_3 : (x, y) \mapsto \left( \frac{\sqrt{3} x^2 - 2 x + \sqrt{3} y^2 - \sqrt{3}}{3x^2 + 2 \sqrt{3} x + 3 y^2 + 1} , \frac{4 y}{3x^2 + 2 \sqrt{3} x + 3 y^2 + 1} \right)$$ of period $3$.