Mapping on a set with respect to function composition

In Isaacs' Algebra, I found the following exercise

Let $G$ be a group of mappings on a set $X$ with respect to function composition. Find an example where $G$ is not a subset of $\operatorname{Sym}(X)$ and $|G|\geq 2$.

I don't understand what "mapping on a set $X$ with respect to function composition" means.

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The thing you did not understand is "group of mappings on a set $X$ with respect to function composition". Not as you quoted! I guess you know what a function and function composition is... And, mapping is nothing but a function; it takes some element of $X$ and just sends it to another one. Further, to understand the whole question, you need the group definition, which I am not sure you already know. – Metin Y. May 24 '13 at 4:31
Okay, I understand what you said. What does "with respect to function composition" mean here? – Paul S. May 24 '13 at 4:33
You know $(\mathbb Z, +)$ is a group? In short, there is a binary operation "$+$" doing $3+4=7$, $8+11=19$ etc. In your example, the group consists of some functions(mappings) and the binary operation now is "$\circ$", i.e, it does $f\circ g$, $h\circ j$ etc. If you wish to see the definition of function composition, see here: en.wikipedia.org/wiki/Function_composition – Metin Y. May 24 '13 at 4:38
Okay, now I'm confused what could be in $G$ that's not in $\text{Sym}(X)$. Every element needs an inverse, so shouldn't every element be in $\text{Sym}(X)$? – Paul S. May 24 '13 at 4:55
Knowing that $Sym(X)$ consists of bijections... What about a function $f \in G$ such that it maps all elements of $X$ to a single element? – Metin Y. May 24 '13 at 5:00