tensors - Reduce the output from tuples by including symmetry?

I need all the possible 3x3 binary tensors, but I'd like to have this account for symmetries. I've started by using the Tuples command.

Tuples[{1, 0}, {3, 3}]

This tuples command produces 512 binary tensors. If I consider tensors that are symmetric by any permutation to be redundant, how can I remove the extraneous ones from the output? Here are a few examples of tensors that I'd like to consider as the same:

{{0, 1, 0}, {0, 1, 0}, {1, 1, 0}} // MatrixForm
{{1, 1, 0}, {0, 1, 0}, {0, 1, 0}} // MatrixForm
{{0, 1, 0}, {0, 1, 0}, {0, 1, 1}} // MatrixForm
{{0, 0, 1}, {1, 1, 1}, {0, 0, 0}} // MatrixForm

Additionally, I'd like to remove tensors that are equal if you replace the 1's with 0's, and vice versa, such that these two tensors are equivalent:

{{1, 0, 1}, {1, 0, 1}, {0, 1, 0}} // MatrixForm
{{0, 1, 0}, {0, 1, 0}, {1, 0, 1}} // MatrixForm

By my estimation, this should reduce the output from 512 to ~50 tensors.

Any help would be greatly appreciated.

Clarification of "permutations" for the original question:

I'm not sure the most accurate way to describe it. Imagine taking a 3x3 array of 1's and iteratively changing each 1 to a 0 until you have an array of 0's. I consider a duplicate being an array that is identical under any rotation or mirror around any row or diagonal.

For example, if you start with:

{{1, 1, 1}, {1, 1, 1}, {1, 1, 1}} 

There are only three distinct 1's that can be removed:

{{0, 1, 1}, {1, 1, 1}, {1, 1, 1}}
{{1, 0, 1}, {1, 1, 1}, {1, 1, 1}} 
{{1, 1, 1}, {1, 0, 1}, {1, 1, 1}}

Answer

First, get set t2 by removing those tensors that are permutations. Second, get set t3 by removing the tensors that are equivalent when the 0-1 substitutions are considered.

t = Tuples[{1, 0}, {3, 3}];
t2 = DeleteDuplicates[t, Sort@#1 == Sort@#2 &];
t3 = DeleteDuplicates[t2, Sort@(#1 /. {0 -> 1, 1 -> 0}) == Sort@#2 &];
Length@t3

This gives a list with 60 tensors.

Edit:

Given the enhanced explanation about the permitted transformations, I think the following approach may give some idea on how to remove the redundant tensors. Here, I consider the rules that check the mirroring. In its present form, this solution gives 212 matrices (in t2), the number of which can be further reduced by including additional conditions for equivalent tensors (like the 0-1 substitutions). If you also consider tensors being equivalent after multiple mirroring operations, etc., those have to be also implemented. Although this approach seems quite tedious, I hope it may still be satisfactory.

t = Tuples[{1, 0}, {3, 3}];
t2 = DeleteDuplicates[t,
     #1 == Reverse@#2(*mirror with respect to the 2nd row*)||
     #1 == Reverse[#2, 2](*mirror with respect to the 2nd column*)||
     #1 == Transpose@#2(*diagonal*)||
     #1 == Reverse@Transpose@Reverse@#2(*the other diagonal*)
     &];

Edit 2:

Ok, so let me try to complete this answer. The solution by @dpholmes provides the answer to the first part of the question. Here, I will continue from that (list t2) and show how to take into account also the 0-1 substitutions. This results in 51 distinct arrays (t3).

t = Tuples[{1, 0}, {3, 3}];
t2 = DeleteDuplicates[t,
     #1 == Reverse@#2(*mirror with respect to the 2nd row*)||
     #1 == Reverse[#2, 2](*mirror with respect to the 2nd column*)||
     #1 == Transpose@#2(*diagonal*)||
     #1 == Reverse@Transpose@Reverse@#2(*the other diagonal*)|| 
     #1 == Reverse@Transpose@Reverse@Reverse[#2, 1]
     (*mirror with respect to the 2nd row and first diagonal*)||
     #1 == Reverse@Transpose@Reverse@Reverse[#2, 2]
     (*mirror with respect to the 2nd row and other diagonal*)||

     #1 == Reverse[Reverse[#2], 2]
     (*mirror with respect to the 2nd row and the 2nd column*)
     &];
t3 = DeleteDuplicates[t2,
     (#1 /. {1 -> 0, 0 -> 1}) == Reverse@#2(*mirror with respect to the 2nd row*)||
     (#1 /. {1 -> 0, 0 -> 1}) == Reverse[#2, 2](*mirror with respect to the 2nd column*)||
     (#1 /. {1 -> 0, 0 -> 1}) == Transpose@#2(*diagonal*)||
     (#1 /. {1 -> 0, 0 -> 1}) == Reverse@Transpose@Reverse@#2(*the other diagonal*)|| 
     (#1 /. {1 -> 0, 0 -> 1}) == Reverse@Transpose@Reverse@Reverse[#2, 1]
     (*mirror with respect to the 2nd row and first diagonal*)||

     (#1 /. {1 -> 0, 0 -> 1}) == Reverse@Transpose@Reverse@Reverse[#2, 2]
     (*mirror with respect to the 2nd row and other diagonal*)||
     (#1 /. {1 -> 0, 0 -> 1}) == Reverse[Reverse[#2], 2]
     (*mirror with respect to the 2nd row and the 2nd column*)
     &];