I'm sorry for the silly doubt.
Assume that $(Epi,Mono)$ is a factorization system on $\mathfrak{C}$. Let $f$ be a monomorphism and consider its factorization $f=m \circ e$. Is it true that by the uniqueness (up to isomorphisms) of the factorization, this is the same as $f=f \circ 1$ (in other terms $m$ is equivalent to $f$ and $e$ to $1$)?
Does the previous result also show that whenever we have an $(Epi,Mono)$-factorization $f=m \circ e$, actually $m$ is an extremal monomorphism (using the diagonalization property!)? This would imply that any category with an $(Epi,Mono)$ factorization system on it has the property that monomorphisms and extremal monomorphisms agree.
Yes, it looks correct. In any such category also all epimorphisms will be extremal. For if $e:X\to Y$ is epi and $m:S\to Y$ is a monomorphism such that $e$ factors through $m$ as $\overline e$, then both $id_Y\circ e$ and $m\circ \overline e$ are epi-mono factorizations of the same morphism, which means that there is a comparison isomorphism $\theta:S \to Y$ such that $\theta \circ m = id_Y$, hence $m$ is invertible. You have shown in your question that a similar argument shows that each monomorphism $m$ is extremal.