The definition from isomorphism
$f$ and $g$ are isomorphisms iff $f.g=id$ and $g.f=id$
Well
$id.id=id$
Does this make $id$ an isomorphism?
My intuition says that not, because that would break the terminal object definition, right?
Terminal object $t$ is a object that has only and only one morphism $f_{t}$ going from any object to it.
I understood that if I have to terminals $t_1$ and $t_2$ there must to be the isomorfisms $f_{t1} : t1 \rightarrow t2$ and $f_{t2} : t2 \rightarrow t1$ and that $f_{t1}.f_{t2}=id_{t1}$
But if I take $id_{t1}$ as an isomorphism, and aslso $f_{t2}$ then there would be two isomorphisms from "any object" to $t1$ and that would break the unique up to unique isomorphisms don't?
Or that means that $f_{t1}$, $f_{t2}$, $id_{t1}$, $id_{t2}$ are all the same morphism? (That would make sense for poset)
$id$ is an isomorphism, yes.
In your case, you have to be careful about domains and codomains. For instance, $f_{t_2}$ is the unique morphism from $t_2$ to $t_1$, if $t_1$ is terminal. And similarly, $id_{t_1}$ is the unique morphism from $t_1$ to $t_1$, if $t_1$ is terminal.
Maybe recalling the definition of terminal object can help : $x$ is terminal if and only if, for each $y$ in the category, there is a unique morphism from $y$ to $x$. So, this unique morphism could be written $f_y$, since it depends only on $y$ (assuming $x$ is fixed).