Is this an equivalence relation on $A$?

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Let $S$ be a non-empty set, $A$ be the set of all functions from $S$ to $S$, and $B$ be the set of all bijective functions from $S$ to $S$. Define the following relation on $A$:

$$f \sim g \quad :\iff\quad \exists h \in B :\quad f\circ h = g.$$

Is this an equivalence relation on $A$?

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Reflexivity

Does $f \sim f$ ? Let $\mathbf 1$ be the identity function. Obviously it is bijective, hence $\mathbf 1 \in B$. Therefore $f = f \circ \mathbf 1$.

Symmetry

If $f \sim g$, does $g \sim f$ ? To prove it, remember that if $h : S \to S$ is a bijective function, it defines an inverse function $h^{-1} : S \to S$ such that $h \circ h^{-1} = \mathbf 1$.

Transitivity

If $f \sim g$ and $g \sim e$, does $f \sim e$ ? To prove it, remember that if $h$ and $k$ are two bijective functions, so is $h \circ k$.