Exponential of Operators Let $H$ be an Hilbert Space $\exp(T)$ the exponential for an operator $T \in L(H)$.
I know that $\exp(A)^{*} \exp(A)=\exp(A) \exp(A)^{*}=id$. Can I conclude that $A^{*}A=AA^{*}$?
Cannot find an counter example neither succeeded my attempts at rearranging. Any hint is appreciated.
EDIT:
For my task i just need that i can conclude
$$ \exp(A)\exp(A)^{*} = \exp(A)^{*}\exp(A) = \exp(A+A^*) $$
 A: The result that most resembles what you might be looking for is the following result of E. Hille (see Theorem 4 of Page 54 here) : For an operator $T \in L(H)$, we say that $\sigma(T)$ is incongruent $\pmod{2\pi i}$ if
$$
\sigma(T)\cap [\sigma(T) + 2k \pi i] = \emptyset\quad\forall k\in \mathbb{Z}\setminus \{0\}
$$
In other words, if $\lambda_1,\lambda_2 \in \sigma(T)$ such that $\exists k \in \mathbb{Z}$ such that $2k\pi i = \lambda_1 - \lambda_2$, then $\lambda_1 = \lambda_2$.
The theorem then states that

If $T_1, T_2 \in L(H)$ are such that $e^{T_1} = e^{T_2}$ and $\sigma(T_1)$ is incongruent $\pmod{2\pi i}$, then $T_1$ and $T_2$ commute.

Assuming this result, we can prove 

If $A \in L(H)$ is such that $\sigma(A)$ is incongruent $\pmod{2\pi i}$ and $e^A$ is a unitary, then $A$ is normal.

Proof:
Simply consider
$$
e^{A^{\ast}} = (e^A)^{\ast} = (e^A)^{-1} = e^{-A}
$$
So by Hille's result, $A^{\ast}$ and $-A$ commute.

I am not sure how to check that the spectrum is incongruent $\pmod{2\pi i}$, but one simple condition could be controlling the spectral radius (for instance if $r(A) \leq \pi$ and $i\pi \notin \sigma(A)$)
