Why are NPI problems not all of the same complexity?
You cannot show that a problem is $\mathsf{NPI}$ without separating $\mathsf{P}$ from $\mathsf{NP}$. 

There are artificial problems that can be proven to be in $\mathsf{NPI}$ using generalizations of Lander's theorem (also see this) assuming that $\mathsf{NP}\neq\mathsf{P}$. 

Also padded version of $ \mathsf{NEXP\text{-}complete} $ problems are $\mathsf{NPI}$ assuming $ \mathsf{NEXP} \neq \mathsf{EXP} $ (see also this and this).

A problem in $\mathsf{NP}$ is often conjectured to be $\mathsf{NPI}$ when:


we can show under reasonable assumptions that it is not $\mathsf{NPC}$ yet it is not known to be in $\mathsf{P}$,
we can show under reasonable assumptions that it is not in $\mathsf{P}$ yet it is not known to be in $\mathsf{NPC}$,


and sometime just when we cannot show that it is in $\mathsf{NPC}$ or $\mathsf{P}$.

An example of a reasonable assumption is the exponential time hypothesis (or some of other computational hardness assumptions).


  Basically what I am asking is why would a problem that can be satisfied in polynomial time (if at all) but not solved in polynomial time (as long as P doesn't equal NP) not be polynomial time reducible to each other. 


I don't see why one would expect that to be true. But in any case, assuming $\mathsf{NPC}\not \subseteq \mathsf{P}$ it follows from Lander's theorem that there are infinitely many levels of $\mathsf{P}$-degrees between $\mathsf{P}$ and $\mathsf{NP}$.