Operators $A$ such that $e^A$ is norm preserving Let $X$ be a Banach space. $A$ a bounded operator. We can define the exponential of $A$ by 
$$e^{A}=\sum_{n=0}^{+\infty}\frac{A^n}{n!},$$
which is also a bounded operator.
Is there any sufficient condition on the operator $A$ to ensure that $e^{A}$ is a norm preserving operator (isometry) i.e. $|e^Ax|=|x|$ for all $x\in X$ ?
If $X=\mathbb{C}$, we get this if $A=a\in i\mathbb{R}$. Maybe if my intuition is right, we get also this in the case  $X=\mathbb{R^n}$ and $A$ has all its eigenvalues purely imaginary. 

The natural generalization of what you noticed in $\Bbb C$ is the condition $A^* = -A$ in any Hilbert space. 
