Definition:
Let $X \subseteq \mathbb R^2.$ A symmetry of $X$ is an isometry $f: \mathbb R^2 \to \mathbb R^2$ such that $f(X) = X.$
Statement:
Let $S = \{(x, y) \in \mathbb R^2 : -1 \le x, y \le 1\}$ be square in the plane. Then $R_{90}(a, b) = (-b, a)$ is a symmetry of $S.$
I will try to prove the statement above using the given definition.
The rotation by $90$ deg is given by $f: \mathbb R^2 \to \mathbb R^2$ as $f(a, b) = (-b, a)$. Since $d((a, b), (c, d)) = \sqrt{(a – c)^2 + (b – d)^2}$ and $d(f(a, b), f(c, d)) = d((-b, a), (-d, c)) = \sqrt{(-b + d)^2 + (a – c)^2}$, we see that $f$ is an isometry.
Let $X$ be a subset of $\mathbb R^2$ . Suppose $y = (-b, a) \in f(X)$. Then $-1 \le -b, a \le 1$ and so $y \in X$. Also, suppose $z = (a, b) \in X$. Then $-1 \le a, b \le 1$ and so $z \in f(X)$. So, $f(X) = X$. Then $f$ is a symmetry of $X$.
Does that make sense?