Let $q=p^n$, for some prime $p$ and positive integer $n$. Let $m$ also be a positive integer, and denote $Tr$ to be the trace of $GF(q^m)$ over $GF(q)$.
I have the suspicion that all the functions $f(x): GF(q^m)\rightarrow GF(q^m)$ which preserve the Kernel of the trace, i.e.
$$ Tr(x)=0 \Leftrightarrow Tr(f(x))=0, \;\text{ for all } x\in GF(q^m), $$
are of the form $f(x)=cx^{p^r}$ for $c\in GF(q)$, $c\neq 0$, and $0\leq r < mn$.
It is not hard to show that functions of the above form do preserve the trace, however I was not able to show the other direction.
Is my suspicion true? If yes, how can we show that every functions that preserves the trace is of the form given above?
I don't see how this could be true. There are just too many functions preserving the kernel of the trace.
For example consider the case $q=2$, $m=5$. The kernel of the trace has $16$ elements, and $16$ other elements in $GF(32)$. The function $f$ can permute both sets of $16$ any which way you want. This gives us $(16!)^2$ bijective functions that preserve the kernel. OTOH there are only ten functions of the form $cx^{2^r}$, $c\in GF(2), 0\le r<5$.
Observe also that all those functions can be realized as polynomials by applying Lagrange's interpolation formula.
This was too easy, so I suspect that you have left out a key assumption.
Guessing that you want the function $f$ to be a bijection that is linear over $GF(q)$. Let's do the same counting exercise in this case as well. Fix a basis $B=\{x_1,x_2,\ldots,x_m\}$ for $GF(q^m)$ over $GF(q)$ such that the first $m-1$ vectors span the kernel of the trace. A linear mapping $f$ preserves the kernel of the trace if and only if its matrix with respect to $B$ is of the form $$ M_B(f)=\left(\begin{array}{ccccc} f_{11}&f_{12}&\cdots&f_{1,m-1}&f_{1m}\\ f_{21}&f_{22}&\cdots&f_{2,m-1}&f_{2m}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ f_{m-1,1}&f_{m-1,2}&\cdots&f_{m-1,m-1}&f_{m-1,m}\\ 0&0&\cdots&0&f_{m,m} \end{array}\right) $$ with $f_{m,m}\neq0$. The mapping $f$ is bijective, iff (in addition to $f_{m,m}\neq0$) the upper left $(m-1)\times (m-1)$ block is an invertible matrix.
It is clear that in all (except possibly some small degenerate cases) such matrices outnumber the scalar times a power of Frobenius mappings that you listed.
There is just so much wiggle room inside the kernel for your list to be anywhere near complete.