The norm is at least $|2+3i|$ by definition, so it is not unitary which implies norm $1$. It can't be selfadjoint, because the eigenvalues of selfadjoint operators must be real. It could be normal though, because $A^*$ being the scalar times the identity is suitable.
Let $T$ be normal. By the spectral theorem, there is a unitary map $V : H \to L^2(\mu)$ for a suitable measure $\mu$ on a measure space $X$ such that we have
$$
T = V^\ast M_f V
$$
for some bounded function $f : X \to \Bbb{C}$, where $M_f : L^2(\mu) \to L^2 (\mu), g \mapsto f\cdot g$ is the multiplication operator which multiplies by $f$.
By the usual properties of the spectral calculus, we have $\varphi(T) = V^\ast M_{\varphi \circ f} V$ for every bounded measurable $\varphi : \Bbb{C} \to \Bbb{C}$. In particular, we have $|T|= V^\ast M_{|f|} V$.
Now, it is easy to see that there is a measurable $g : X \to \Bbb{C}$ with $|g(x)| = 1$ for all $x \in X$ and $|f(x)| \cdot g(x) = f(x)$. In fact, the function
$$
g : X \to S^1 \subset \Bbb{C}, x \mapsto \begin{cases} \frac{f(x)}{|f(x)|}, & f(x) \neq 0,\\
1, & f(x) = 0\end{cases}
$$
does the job.
Since $g$ has modulus one, the multiplication operator $M_g$ is unitary. Hence, so is $U = V^\ast M_g V$ and we have
$$
T = V^\ast M_{f} V = V^\ast M_g M_{|f|} V = V^\ast M_g VV^\ast M_{|f|} V = U\cdot |T|
$$
as desired.
EDIT: As I wrote in the comments, the unitary $U$ is not unique in general. For example for $T=0$, we have $|T|=0$, so that any unitary $U : H \to H$ will do the job.
Best Answer
Let $U$ be a unitary, then $UU^*=U^*U=1$.
$U^*U=1$ implies that $U$ is an isometry. Also $UU^*=1 $ implies that $U$ is onto, because ($UU^*H=H$).
Conversely, Suppose $U$ is an onto isometry, so $U^*U=1$. It's just necessary to show that $UU^* = 1$. If $\eta\in H$, then there is $\xi\in H$ such that $U\xi=\eta$ ($U $ is onto). Also $U$ is an isometry, so $ \xi=U^*U\xi = U^*\eta$ .Now clearly $UU^*\eta = U\xi =\eta$ for $\eta \in H$. Therefore $UU^*=1$.