I am considering a $3\times 3$ block of pixels where each pixel can be either $0$ or $1$.
How many unique blocks are there if we consider rotations/reflections to be identical?
I believe the answer is $76$, but I am really bad at combinatorics. Below is an image of what I believe is the solution set:
I would classify all patterns in the following way.
I assume that black pixels are meaningful ($1$), the rest is a background ($0$).
First we see that the lower row in your example is just the negative for the upper row so the total number of patterns = $2 \times $ number for the upper row.
So let's concentrate on the upper row.
Here we have cases where in the $ 3\times{3}$ grid there is no pixels, there is 1 black pixel, there are 2 pixels, 3 pixels and 4 pixels. These cases I denote as #0,..#4. In this grid the central pixel is stabile under rotation and reflection, the remaining $8$ ones make structure which I'll name outer cycle (neighborhood of the central pixel) which generally can be affected by these operations.
Evidently $1$ pattern.
Introduction for other cases.
For other cases it is important to notice that we can distinct between patterns with visible central pixel and without it. Patterns with central pixel for #k (denoted as PC) are descended from the lower case #k-1. Patterns without central (denoted as PW) pixel can be classified according to 'distances' between pixels measured on the outer cycle (there are $8$ pixels in this outer cycle, so the sum of number of pixels and distances is always equal to $8$ as it was said). Distances are just empty pixels between elements from the right and left side of the pixel to the nearest pixel - e.g. in the case of two pixels in the outer cycle two numbers can denote a distance.
If a distance is equal to $0$ it means that two pixels are in the one continuous block. These possible distances are listed below in $\{,\}$ brackets for every case. The number of distances in the set $\{,\}$ is equal to the number of pixels in the pattern.
Notice that rotation and reflection doesn't change the set of distances for a given pattern so to use distances for the classification it seems to be a good idea.
PW - we have here distance $\{ 7\} $ - there are two distinct patterns for this set of distances (a pixel in the corner of outer cycle and a pixel in the midpoint of outer cycle) as in the drawing in the question
PC - $1$ pattern
Total number $=2+1=3$ patterns.
PW -
2 pixels so the sum of distances must be $8-2=6$
distances $\{0,6\}$ - $1$ pattern
$\{1,5\}$ - $2$ patterns
$\{2,4\}$ - $2$ patterns
$\{3,3\}$ - $2$ patterns
PC - $2$ patterns ( based on PW#1)
Total number $= 7+2 =9$
PW - 3 pixels so the sum of distances must be $8-3=5$
distances $\{0,0,5\}$ - $2$ patterns
$\{1,1,3 \}$ - $2$ patterns
$ \{1,2,2 \}$ - $2$ patterns
$\{1,0,4 \}$ - $2$ patterns
$ \{3,0,2 \}$ - $2$ patterns
PC - $7$ patterns ( based on PW#2)
Total number $= 10+7=17$
PW
4 pixels so the sum of distances must be $8-4=4$
distances $\{4,0,0,0\}$ $1$ pattern
$\{3,0,0,1\}$ $4$ patterns - the only situation for all cases when for a given set of distances we have $4$ patterns not $2$ or $1$.
$\{2,0,0,2\}$ $2$ patterns
$\{2,0,1,1\}$ $2$ patterns
$\{1,1,1, 1 \}$ $2$ patterns
PC $10$ patterns ( based on PW#3)
Total number $=11+10= 21$
Summary
Total sum for the upper row: $1+3+9+17+21=51$
And multiplying by $2$ we receive $2 \times {51} = 102$ patterns.
It's not the same number as you have given, but your visualization of $76$ patterns was very helpful to construct this solution.
