Solution: Let us see a general result along the same line. Let $G$ be a cyclic group of order $n$. Let $g$ be a generator of $G$. Then \[ o\left( g^k \right) = \frac{n}{\gcd (n,k)}. \] Therefore, $g^k$ will be another generator of $G$ if $o\left( g^k \right) = n$, that is $\gcd (n,k) = 1$. Therefore, the number of generators of a cyclic group of order $n$ is \[ \#\left\{ k \in \mathbb{N} : \gcd (n,k) = 1, \text{ and } k \leq n \right\} = \phi (n), \] where $\phi$ is the Euler-phi function.

The number of generators of a cyclic group of order $36$ is \[ \phi (36) = \phi (2^2 \times 3^2) = \phi (2^2) \times \phi (3^2) = 2 \times 6 = 12. \]