Can every cardinal number between $\kappa^+$ and $2^\kappa$ be realized in this way? (Assume ZFC for the entire question. By a tree, I mean a tree in the sense of set theory. I write $h(T)$ for the height of a tree $T$, and $h(f)$ for the height of an element $f \in T$.)
Definition 0. Whenever $X$ is a set, write $X^{inj}$ for the collection of all injections $f$ such that:


*

*$f$ has codomain $X$

*There exists an ordinal $\alpha$ such that $f$ has domain $\alpha$.


Proposition 0. $|X^{inj}|=2^{|X|}$ (See here,  Asaf Karagila's answer).
Now observe, furthermore, that $X^{inj}$ is a tree under the restriction order. Explicitly, this is the order $\leq$ defined on $X^{inj}$ by asserting that $f \leq g$ iff $f$ is a restriction of $g$.
It follows that for all $f \in X^{inj}$, the height of $f$ in $X^{inj}$ is just
$\mathrm{dom}(f)$. Hence:
Proposition 1. $|h(X^{inj})| = |X|^+$.
As it turns out, we can view $|X^{inj}|$ and $|h(X^{inj})|$ as two ends of a spectrum. To see this, let us firstly define the concept of a coherent equivalence relation on a tree.
Definition 1. Let $T$ denote a tree and let $\sim$ denote an equivalence relation on $T$. Then $\sim$ is coherent iff $f \sim g$ implies $h(f)=h(g)$, for all $f,g \in T.$ 
It follows that every tree has a least coherent equivalence relation, namely the equality relation, and a greatest coherent equivalent relation, namely the relation $\sim$ such that $f \sim g$ iff $h(f)=h(g)$. The above two results can thereby be rehashed as follows.
Proposition 2. Let $X$ denote a set, $A$ denote the least coherent equivalence relation on $X^{inj}$ (namely equality) and $B$ denote the greatest coherent equivalence relation on $X^{inj}$ (namely level-by-level equivalence). Then:


*

*$|X^{inj}/A| = 2^{|X|}$

*$|X^{inj}/B| = |X|^+$


The question is whether or not every cardinal number between $2^{|X|}$ and $|X|^+$ is realized as $|X^{inj}/\sim|$ for some coherent equivalence relation $\sim$ on $X^{inj}$.

Question. Let $X$ denote a set and $\kappa$ denote a cardinal number between $|X|^+$ and $2^{|X|}$. Does there necessarily exist a
  coherent equivalence relation $\sim$ on $X^{inj}$ such that
  $|X^{inj}/\sim| = \kappa$?

Extra credit: if the answer is yes, can we always choose $\sim$ such that $X^{inj}/\sim$ is itself a tree?
 A: Your definition of "coherent" is extremely weak: a coherent equivalence relation is really just a separate equivalence relation on each level of $X^{inj}$.  You can choose these equivalence relations on each level freely to have whatever number of equivalence classes you want (with a minimum of $1$ and a maximum of the cardinality of the level), so then the total number of equivalence classes will just be the sum of the number on each level.  It is very easy to arrange this so that the total number is any cardinal between the obvious bounds.
Here's one explicit way to do it.  Let's write $\lambda=|X|$ and suppose $\lambda^+\leq\kappa\leq 2^\lambda$.  Note that the $\lambda$th level of $X^{inj}$ has cardinality $2^\lambda$, so we can pick an equivalence relation $\sim_0$ on the $\lambda$th level with $\kappa$ equivalence classes.  Now define $f\sim g$ if either $h(f)=h(g)<\lambda$ or $h(f)=h(g)\geq\lambda$ and $f|_\lambda\sim_0 g|_\lambda$.  This equivalence relation has one equivalence class for each level below $\lambda$ and then has $\kappa$ equivalence classes for each level $\lambda$ and above, giving $\lambda+\lambda^+\cdot\kappa=\kappa$ total equivalence classes.  The quotient by this equivalence relation is naturally a tree, since $f\sim g$ implies $f|_\alpha\sim g|_\alpha$ for any $\alpha$.
