Type of isometry What type of isometry of $\mathbb{R}^3$ is the one given by sending $(x,y,z)$ to $(y,z,x)$ for each $x,y,z\in\mathbb{R}$? How can I find this isometry? Does anyone have a hint?
My solution is that first from $(x,y,z)$ to $(z,y,x)$, reflection by $z=x$, then from $(z,y,x)$ to $(y,z,x)$, reflection by $x=y$? Is that right?
 A: Here are some hints to help you work it out. First note that the isometry $f\colon\mathbb{R}^3\rightarrow\mathbb{R}^3$ defined by $(x,y,z)\mapsto(y,z,x)$ has the matrix $$M=\left(\begin{array}{ccc}
0&1&0\\
0&0&1\\
1&0&0
\end{array}\right)$$
associated to it (by left multiplication)*. Can you see what the determinant of this matrix is? What does it mean for $f$ if the determinant of $M$ is that value? Can it be a reflection? How about a rotation? (specifically, can we tell if $f$ preserves orientation?)
Note that $f(0,0,0)=(0,0,0)$. Can $f$ be a nontrivial translation or glide rotation/reflection if this is the case?
What are the only other options available?
*If you're not quite sure how I got this matrix, please feel free to ask.
A: First, consider the same situation in two dimensions. Observe that any vector $(x,y)$ is mapped to $(-y,x)$ by a rotation of $90^{\circ}$ (analogous to finding the hat vector). Observe also that reflection across the $y$-axis is simply a change of sign of the $x$-component. So $(x,y) \to (y,x)$ by a rotation of $90^{\circ}$ and a reflection in the $y$-axis. Let's try and generalize this to 3D.
Consider any vector $(x,y,z)$ in three dimensions. When performing the rotations we can think of the vector as not having a third dimension and simply find the hat vector in 2D as we did above. In order to obtain $(x,y,z) \to (y,z,x)$ we want $x \leftrightarrow y$ followed by $x \leftrightarrow z$, i.e. $(x,y,z) \to (y,x,z) \to (y,z,x)$. Let $\theta$ be the polar angle (angle down from $z$-axis) and $\phi$ the azimuthal angle. Then by a rotation of $90^{\circ}$ in the azimuthal angle and a reflection in the $yz$-plane gives us $(y,x,z)$. A rotation of -$90^{\circ}$ in the polar angle and a reflection in the $xz$-plane gives us $(y,z,x)$.
So rotation $\to$ reflection $\to$ rotation $\to$ reflection certainly does it. Perhaps there is a way to prove what can and cannot be done. I think the matrix approach as suggested by Daniel is the way to go with that.
