I think this is wrong, but close. Consider the mapping of a pole to the polar wrt an arbitrary fixed circle with center $O$. Then it is easy to show that for any point $P$ on a line $l$, then polar of $P$ passes through the pole of $l$. It is easy to show for injectivity and surjectivity, but the only problem is that it is not really surjective since there is no point that maps to a line passing through $O$, and $f(O)$ is not defined.

Suppose, $ \Lambda_2 $ is the image of the plane $ \Lambda_1 $ under the transformation. Consider the pre-images of a set of parallel lines. From the given a condition they are collinear on some line $ l $. Now consider two different lines $ l_1 $ and $ l_2 $ on $ \Lambda_1 $ such that the image of the points on $ l_1 $ are parallel and similar for $ l_2 $. If $ l_1 $ and $ l_2 $ intersect at some point $ X $ then the image of $ X $ will have to be parallel to the images of the points on the line $ l_1$ and also to the images of the points on the line $ l_2$. But clearly its impossible. So $ l_1 $ and $ l_2$ are parallel. So any line for which the points on it will have parallel images are parallel to this two lines. Now take two parallel lines $ k_1$ and $ k_2$ which are not parallel to $ l_1 $. So the images of the points on $ k_1$ concurr at some point $ X_1$ and similarly define $ X_2$. Now consider the line $X_1X_2$. If its pre-image is $ X$ then $X$ has to lie on $k_1$ and also on $k_2$ which is clearly impossible since $k_1\parallel k_2$. So no such transformation exists.