Q:

Finding all possible combination **patterns** - as opposed to all possible combinations

I start off with trying to find the number of possible combinations for a 5x5 grid (25 spaces), where each space could be a color from 1-4 (so 1, 2, 3, or 4)
I do 4^25 = 1,125,899,906,842,624 different combinations
However, now I'm trying to change the number of combinations to account for grids with the same number pattern, for example:
{ 1 1 1 1 1 }
{ 3 3 3 3 3 }
{ 4 4 2 2 3 }
{ 4 3 2 1 1 }
{ 2 2 1 2 3 }
1 is now 2, 2 is now 4, 3 is now 1, 4 is now 3

{ 2 2 2 2 2 }
{ 1 1 1 1 1 }
{ 3 3 4 4 1 }
{ 3 1 4 2 2 }
{ 4 4 2 4 1 }
I'm having trouble trying to come up with an equation I can use to solve this for a (x * y) grid where each space could be a color from 1 to (c).

A:

You need to split the count according to the number of different numbers used. There are $4$ grids using one number each, but they all yield the same pattern.
At the other extreme there are
$$4^{25}-\binom43\cdot3^{25}+\binom42\cdot2^{25}-\binom41\cdot1^{25}=4^{25}-4\cdot3^{25}+6\cdot2^{25}-4$$
grids that use all four numbers. The four numbers can be permuted in $4!=24$ different ways, so there are
$$\frac1{24}\left(4^{25}-4\cdot3^{25}+6\cdot2^{25}-4\right)$$
patterns.
Given two colors, say $a$ and $b$, we can fill them into the grid in $2^{25}-2$ ways (since we need to exclude the one-number grids). However, that counts each pattern twice, since we can interchange $a$ and $b$, so this case contributes $2^{24}-1$ patterns.
Similarly, three colors can be filled into the grid in
$$3^{25}-\binom32\cdot2^{25}+\binom31\cdot1^{25}=3^{25}-3\cdot2^{25}+3$$
ways, but the colors can be permuted in $3!=6$ ways, so the number of patterns is only
$$\frac16\left(3^{25}-3\cdot2^{25}+3\right)=\frac12\left(3^{24}-2^{25}+1\right)\;.$$

A:

$\sum _{n=1}^c \mathcal{S}_{x y}^{(n)}$ is what you're after, with $x, y, c$ the two dimensions and the number of colors. $\mathcal{S}_{x y}^{(n)}$ is the Stirling number of the second kind. 
This generalizes to arbitrary dimensions.

