Isometries and Preservation of Eigenvalues Does conjugation by an isometry preserve eigenvalues? If not, are there certain (non-trivial) situations where it does? 
 A: Conjugation by anything (invertible) preserves the characteristic polynomial, and therefore preserves the eigenvalues (see this thread for example). For completeness here's the argument:
Let $A$ be any matrix, and let $Q$ be any invertible matrix (need not be an isometry). Then
$$\begin{align*}
\text{characteristic polynomial of }A&=\det(A-\lambda I)\\\\
&=\det(Q)\det(A-\lambda I)\det(Q)^{-1}\\\\
&=\det(Q)\det(A-\lambda I)\det(Q^{-1})\\\\
&=\det(Q(A-\lambda I)Q^{-1})\\\\
&=\det(QAQ^{-1}-Q\lambda IQ^{-1})\\\\
&=\det(QAQ^{-1}-\lambda I)\\\\
&=\text{characteristic polynomial of }QAQ^{-1}
\end{align*}$$
A: conjugacy preserves the eigenvalues.
here is another way to see that without the use of determinants or characteristic polynomials.
suppose $QQ^\top = Q^\top Q = I, QAQ^{-1} = B.$ let $\lambda$ be an eigenvalue of $A$ and $x\neq 0$ a corresponding eigenvector.  then we have 
$$Ax = \lambda x\implies BQx = QAQ^{-1}Qx=\lambda Qx, Qx \neq 0 $$
that shows if $\lambda$ is an eigenvalue of $A,$ then $\lambda$ is an eigenvalue of $B.$
A: I am aware this question is very old. However, the answers are quite misleading, because they assume that the isometry should be invertible. I want to give a simple counterexample for the case of non-invertible isometries.
We look at the isometry $$I=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}:\mathbb R\to\mathbb R^2.$$ Then, let us consider any diagonal operator on $\mathbb R^2$, i.e., $$A=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ which obviously has the eigenvalues $a$ and $b$. The conjugate operator on $\mathbb R$ now is the operator
$$ I^*AI = \frac 12 \begin{pmatrix}1 & 1\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} = \frac 12(a+b).  $$
As this counterexample shows, what would be needed to preserve an eigenvalue is that its eigenvectors need to be in the range of the chosen isometry. Hence, an isometry can only preserve the eigenvalues of any operator, if it is a bijective isometry - as in the previous answers.
