Is my application of Burnside's Lemma correct in this combinatorial problem? For a course in Combinatorics (I know very little group theory unfortunately), we've been tasked to use Burnside's Lemma on the following problem:
Suppose you write a 5-digit number on a piece of paper (numbers smaller than 10,000 have zeros added in front of them). If you hold the paper upside down, the numbers 0,1,6,8,9 are read as 0,1,9,8,6 respectively. We identify the numbers that are readable both ways (e.g. 89166 and 99168).
The question is of course: how many different numbers are there really?
The problem I have with this exercise is that my calculation differs wildly from that of my fellow students (who find about 50800 different numbers), so I'd like to know what is (possibly) wrong with my calculation, and in the process learn to understand Burnside's Lemma better.
My calculation is as follows:
Out of the 10000 possible numbers, only those numbers wich are legible upside down are of interest to us for now $|S| = 5^5 = 3125$. The permutationgroup consists of the identity, and the 'holding the paper upside down'-permutation.
For the identity, every element is invariant, so we have 3125 elements.
For the other permutation, an element is invariant if it is the same when it is read upside down, so that gives $5^2 \cdot 3 = 75$ possibilities. 
Applying Burnside's Lemma we get $\frac{3125 + 75}{2} = 1600$ unique elements. This seems quite logical to me: about half of the 3125 numbers are actually counted double, however 75 are not, so that gives $\frac{3125 - 75}{2} + 75$ unique numbers.
Finalizing, there are 10000 numbers, subtracting the 3125 numbers that can be read upside down, and adding the unique ones gives 98475 actually unique numbers.
Evidently, my answer is different, however I'm interested to know what is actually correct. Thank you for any help you can provide!
 A: Using Burnside I get the following result. By inspection we have the following cycle index: 
$$\frac{1}{2}(a_1^5 + a_1 a_2^2).$$
We need to compute the number of digit sequences fixed by these two permutations, the identity and the flip. Furthermore the flip exchanges sixes and nines at the same time as it permutes slots.
The identity fixes all $10^5$ digit sequences. For the flip according to Burnside we must have constant values on the cycles taking into account that a six and a nine placed on a two-cycle are also constant because the flip simultaneously exchanges those digits. We cannot have a six or a nine on the one-cycle because they wouldn't be constant on that cycle. This gives the following four possibilities:
$$8\times 8\times 8 + 8 \times 8\times 2 + 8\times 2\times 8 + 8\times 2\times 2.$$
The first factor in these four products corresponds to the choice for the one-cycle and the next two to the choices for the two two-cycles. These are eight or two, depending on whether we choose a digit that is not six nor nine, or we choose either six and nine or nine and six.
Burnside now yields
$$\frac{1}{2}
\left(10^5 + 8\times 8\times 8 + 8 \times 8\times 2 + 
8\times 2\times 8 + 8\times 2\times 2\right) = 50400.$$
The following code snippet was used to verify the above calculation.

#! /usr/bin/perl -w
#

MAIN: {
    my %seen;

    for(my $n=0; $n<10**5; $n++){
        my $m = sprintf "%05d", $n;
        my $rev = reverse $m;

        $rev =~ s/6/x/g;
        $rev =~ s/9/y/g;

        $rev =~ s/x/9/g;
        $rev =~ s/y/6/g;

        my @l = sort { $a <=> $b } ($m, $rev);
        $seen{$l[0] . "-" . $l[1]} = 1;
    }

    print scalar(keys %seen);
    print "\n";
}

