Let $A=\left\{1,2,\cdots,10\right\}$
Let $f,g:A\to A$. Consider the equivalence relation
$$ fRg \iff \exists h:A\to A. f=h\circ g$$ where $h$ is invertible.
Now, let $g(x)=5$:
Why is $\left| \left\{ f\in A\to A : fRg \right\} \right| = 1$?
Update:
If $f(x) = 5$ then it's true that $f = h\circ g$ if $h$ is the identity function.
Now, Why any other function won't work?
I hope that you are defining the relation by $f\sim g \Rightarrow \exists h | f = g \circ h$.
If so, take a $f = g \circ h$, then $im f \subseteq im g = \{ 5 \} $ so clearly $f$ would be the constant function with value $5$, hence $f=g$.