Let $FF$ be the category of forests, whose objects are strictly partially ordered sets such that for a forest $F$, the set $\{y\in F\mid y<x\}$ is a finite linear order for any $x\in F$, whose cardinality is called the level of $x$, and whose morphisms are the level-preserving and order-preserving functions. The trees form a full subcategory $TT$ of $FF$, where a forest is a tree iff there is exactly one element of level $0$. These are basically acyclic directed graphs (the trees), or disjoint unions of trees (the forests). Now I have proven that by cutting the root of a tree one obtains a functor from TT to FF that is an fully faithful and essentially surjective, and hence defines an equivalence of categories, so these categories are equivalent. But I was wondering: are they also isomorphic? I feel like the answer is no, but I can't prove it. Any thoughts or hints?
For (some of) the comments below: I am not asking whether this functor is an isomorphism, because it clearly isn't, but how to show that there exists no functor that is an isomorphism.
There is only one initial object in $FF$, the empty forest $\varnothing$. However, in $TT$, any singleton $\{x\}$ is an initial object. An isomorphism of categories would restrict to a bijection between the set of initial objects in $FF$ and the set of initial objects in $TT$, which doesn't exist.