I have some questions about normal ordering in quantum field theory: I already read [this][1] very good question with very very good answers and [this][2] other question with other very good answers (I read also [this][3] one and many others, but without understanding much).

For what I understood, *normal ordering* is more a simbolic operation than an operator, so if for example I have
\begin{equation*}
\left[\hat{a},\hat{a}^\dagger\right]
=
1
\end{equation*}
Then I'm not authorized to say that $:\hat{a}\hat{a}^\dagger:=:\hat{a}^\dagger\hat{a}:+:1:$ (where I think that $:1:=1$ demonstrated through unitary operators). What I don't understand here is 

 - The main fact is that this operation is non-linear? So (even if [here][1] the answer from Sebastiano Peotta seems to say the opposite)
\begin{equation*}
:\hat{a}^\dagger\hat{a}:+:1:
\neq
:\hat{a}^\dagger\hat{a}+1:
\,?
\end{equation*}

- Or the main fact is that this operation doesn't care about operatorial equalities? In that case I would just have 
\begin{equation*}
:\hat{a}\hat{a}^\dagger:
\neq
:\hat{a}^\dagger\hat{a}+1:
\,?
\end{equation*}

At the same time I wasn't able to find this kind of question, that is the main doubt that I have: 

- what if I rename the operator with the following substitution $\hat{b}^\dagger=\hat{a}$? In that case $:\hat{a}\hat{a}^\dagger:=\hat{a}^\dagger\hat{a}$, but $:\hat{b}^\dagger\hat{b}:=\hat{b}^\dagger\hat{b}=\hat{a}\hat{a}^\dagger$!

Is this crazy or am I doing something wrong (very likely)? 

  [1]: https://physics.stackexchange.com/q/345898/
  [2]: https://physics.stackexchange.com/q/309410/
  [3]: https://physics.stackexchange.com/q/24157/