Let $X \subseteq R^2$. A symmetry of $X$ is isometry $f: R^2 \to R^2$ such that $f(X) = X$.
For example, square has $8$ symmetries one of which is $R_{90}(a, b) = (-b, a)$.
Is an element of $X$ is both an image and pre-image of a symmetry? So, then shouldn't $R_{90}(a, b) = (a, b)$? I understand this isn't correct if we look at it geometrically, but I am trying to understand the algebraic gist of it.
In this case, $X$ is the whole square, so for $R_{90}$ to be a symmetry, we only need that the image of $X$ equals $X$, not that every point in $X$ is fixed by $R_{90}$. So while you're right in that $R_{90}(a,b) \neq (a,b)$, it is true that $$ X = \{R_{90}(x) : x \in X\}, $$ and that's all we need.