Graph of a function invariant under the counter-clockwise rotation for $90^\circ$ around the origin > Let $S\subseteq\Bbb R$. Assume a function $f: S\to S$ has these properties: > > With a $90^\circ$ counter-clockwise rotation of its graph around the origin, we get the same graph $\Gamma_f$ again. Prove that the function must be bijective. * * * I've seen the parametric formula: $$x'=x\cos\theta-y\sin\theta$$ $$y'=x\sin\theta+y\cos\theta$$ In my case:$$\theta=\frac{\pi}{2}$$ My first attempt was to find the properties of a function that satisfies this condition: $f(x)=y\in\Gamma_f \wedge f(-y)=x\in\Gamma_f,$ but it didn't help me. I'm not sure if I understand what "the same" means in this task.

**Alert: just a long comment with the purpose of giving a visual insight**

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According to the post itself and what @conditionalMethod has already stated, let $r_O:\Bbb R^2\to\Bbb R^2$ be our rotation for $90^\circ$ around the origin $O(0,0)$, then: $\color{brown}{r_O:(x,f(x))\mapsto (-f(x),x))}$.

Here is a concrete example of such a function:

$$f(x)=\begin{cases}x+1,&x\in(-2,-1)\\\\-x+1,&x\in(-1,0)\\\\-x-1,&x\in(0,1)\\\x+1,&x\in(1,2)\end{cases}$$ so, e.g. $$S=(-2,2)\setminus\\{-1,0,1\\}$$ I also high-lighted a squre with edges of length $a=\sqrt{2}$ which I found interesting. $\Gamma_f$ is black in the picture: ![enter image description here](

Another example is the function $$\begin{cases}x=\cos\left(\alpha+k\frac\pi2\right)\\\y=\sin\left(\alpha+k\frac\pi2\right)\end{cases},\quad\alpha\in\left(\arctan a,\frac\pi2\right),k\in\\{1,2,3,4\\},$$

whose graph consists of the four red arcs on the unit circle:

![enter image description here](