Image of Homomorphisms for Cyclic Groups I'm reading an example for abstract algebra and I would like to clarify some terms. The example is showing the number of homomorphisms from $\mathbb Z $ to $S_3$.
Take a homomorphism $f$ from $\mathbb Z$ to $S_3$. Since $1$ is a generator of $\mathbb Z$, $f$ is completely determined by the image of $1$. Since $1$ has infinite order, there are no restrictions on the image of $1$. Thus each element of $S_3$ is a possible image for $1$. There are $6$ homomorphisms from $\mathbb Z$ to $S_3$.
What does "completely determined by the image of $1$" mean? I thought about it a little bit, and it seems that since $1$ is a generator of $\mathbb Z$, any element in $\mathbb Z$ can be generated from $1^k$. When it says image of $1$, does it mean $f(1^k)$, where $k$ is an integer? In general, what does the "image" of an element under homomorphism mean? I thought for an element $g$ in group $G$, and for a homomorphism $h$, the image of $g$ under $h$ is $h(g)$. How can that determine the entire homomorphism?
Thanks for your help.

It means that if we give the value of $f(1)$ then we have the value of $f(n)$ for all $n\in\Bbb Z$ since
$$f(n)=f(\underbrace{1+\cdots+1}_{n\;\text{times}})=f(1)\circ\cdots \circ f(1)$$
but since there's $6$ elements of $S_3$ then there's $6$  homomorphisms from $\Bbb Z$ to $S_3$.
