You will not be able to find a proof for arbitrary sized grids. Consider the grid {0,1,2} X {0,1}.
We can build a solution with two primitives, which can be reversed and rotated for all possible orientations.
Non-corner to non-corner. initial position (1,0) and target (1,1). The walk (1,0),(2,0),(2,1),(1,1),(1,0),(0,0),(0,1),(1,1) is a satisfying walk.
Corner to non-corner:
Initial position (2,0) and target (1,1). The walk (2,0),(2,1),(1,1),(1,0),(0,0),(0,1),(1,1),(1,0),(0,0),(0,1),(1,1).
(Edit) For the {0,1,2} x {0,1,2} grid there is a length-22 walk from (1,0) to (2,1) which satisfies your conditions. Namely, (1,0),(2,0),(2,1),(1,1),(1,0),(0,0),(0,1),(0,2),(1,2),(1,1),(0,1),(0,2),(1,2),(1,1),(0,1),(0,2),(1,2),(1,1),(0,1),(0,2),(1,2),(2,2),(2,1).
My insight: Both of these solutions use repeated partial traversals of 4-cycles to correct "imbalances" of neighboring cells. I conjecture that there are solutions for a large family of grids, but the start and end positions are necessarily on the perimeter.
For the case of necklaces we have the cycle index
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
We get with $k$ colors $Q_j$ (this is PET)
$$[Q_1^{n/k} \cdots Q_k^{n/k}]
Z\left(C_n; \sum_{j=1}^k Q_j\right)
= \frac{1}{n} \sum_{d|n} \varphi(d)
[Q_1^{n/k} \cdots Q_k^{n/k}]
\left(\sum_{j=1}^k Q_j^d\right)^{n/d}.$$
We see that $d$ must divide $n/k:$
$$\frac{1}{n} \sum_{d|n/k} \varphi(d)
[Q_1^{n/k} \cdots Q_k^{n/k}]
\sum_{m_1+\cdots+m_k = n/d}
{n/d\choose m_1,\ldots, m_k}
\prod_{j=1}^k Q_j^{d m_j}.$$
Here the $m_j$ are non-negative. Now we must have $d m_j = n/k$
so that $m_j = n/k/d$ and we get
$$\frac{1}{n} \sum_{d|n/k} \varphi(d)
\frac{(n/d)!}{(n/k/d)!^k}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n} \sum_{d|n/k} \varphi(n/k/d)
\frac{(kd)!}{d!^k}.}$$
As a sanity check with just one color $k=1$ we get $\frac{1}{n}
\sum_{d|n} \varphi(n/d) = 1$, which is correct. With $k=n$ colors we get
$\frac{1}{n} \sum_{d=1} \varphi(1/d) \frac{(nd)!}{(d!)^n} = \frac{1}{n}
n! = (n-1)!$ which is correct as well. This is OEIS
A208183.
Best Answer
Your requirements for the placement of colors is equivalent to the requirements for a diagonal Latin square, or as it used to be called, a double diagonal Latin square. According to OEIS, there are just $48$ diagonal Latin squares of order $4$, i.e. of size $4$ x $4$.
As @Arthur pointed out, if rotations and reflections are not considered distinct, we must divide this number by $8$. Hence, there are just $6$ distinct solutions.
EDIT
Just made a brute force program to check that I was right, and it gave the answer $12$, not $6$. Manual checking of the program's answer seems to show the program is correct. Here is the program's answer:
None of these appear to be rotations or reflections of each other. What did I miss?
EDIT 2
As @omegadot points out, the solutions above include reflections about the main diagonals. Removing these cases gives the following: