How many orthogonal endomorphisms $f: \mathbb{R}^3\to \mathbb{R}^3$ are there such that $\operatorname{tr}(f)=3$?
Definition: An orthogonal endomorphism $f:\mathbb{R}^3\to \mathbb{R}^3$ is a linear transformation such that for all $x,y\in \mathbb{R}^3$, we have $\langle f(x),f(y)\rangle =\langle x, y\rangle$
Every orthogonal endomorphism can be perceived as an orthogonal matrix $_\mathcal{A} M(f)_\mathcal{A}$ where $\mathcal{A}$ is an orthonormal basis. The orthogonal group $\operatorname{O}(3)$ consists of rotation matrices $\operatorname{SO}(n)$ and rotary reflection matrices in $\operatorname{O}(3)\setminus \operatorname{SO}(3)$. Any rotation in space can be described by choosing a suitable orthonormal basis through a matrix of the form$$R=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix} \quad \text{ where } \varphi\in [0,2\pi)$$ Likewise, the rotary reflection matrices can be described by a matrix of the form$$S=\begin{pmatrix} -1 & 0 & 0 \\ 0 & \cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix} \quad \text{where } \varphi\in [0,2\pi)$$ Now, we have $\operatorname{tr}(R)=1+2\cos(\varphi)\overset{!}=3$ and $\operatorname{tr}(S)=2\cos(\varphi)-1\overset{!}=3$ in $\varphi\in [0,2\pi)$.
The only soultion is $0$ for the first equation. Does this mean there is only one endomorphism?