Number of elements of order $4$ in $\Aut(\ZZ/20)$

Since \(\mathbb Z/20\mathbb Z\) is cyclic, \[ \operatorname{Aut}(\mathbb Z/20\mathbb Z)\cong(\mathbb Z/20\mathbb Z)^\times. \] By the Chinese remainder theorem, \[ (\mathbb Z/20\mathbb Z)^\times \cong(\mathbb Z/4\mathbb Z)^\times\times(\mathbb Z/5\mathbb Z)^\times \cong C_2\times C_4. \] An element \((a,b)\in C_2\times C_4\) has order \(4\) exactly when \(b\) has order \(4\). There are two choices for \(a\) and two elements of order \(4\) in \(C_4\), hence \[ 2\cdot2=4 \] such elements.

Explicitly, they correspond to \[ 3,7,13,17\pmod{20}. \] Therefore the answer is \[ \boxed{4}. \]