Automorphism with no square root Related to Every normal operator on a separable Hilbert space has a square root that commutes with it
Does it exist an automorphism $f$ in a separable $\mathbb C$ Hilbert space, such that $f$ has no square root?
If so, a concrete example would be useful. 

Yes, such operators exist. This was proven by Halmos, Lumer, and Schäffer, Proc. AMS, 4, 1 (1953), 142-149. 
Concretely, given a domain $D\subset\mathbb C$ define 
$$
D^{1/2}=\{\lambda\in\mathbb C:\ \lambda^2\in D\}. 
$$
They proved that the multiplication operator $M_z\in B(L^2(D))$ given by $(M_zf)(z)=zf(z)$ has a square root if and only if $D^{1/2}$ is disconnected. This can be seen to be equivalent to $D$ surrounding (but not containing, obviously) the origin. 
So, if for instance you take any disk that does not contain the origin, say $D=\{\lambda:\ |\lambda-2|<1\}$, then $M_z\in B(L^2(D))$  is invertible and has no square root. 
