Find files that have the same name as the directory

You can easily match file names that have the same name as the containing directory with a regex using a backreference. These are file names of the form what/ever/something/something where the two something parts are identical and don't contain a slash. In BRE regex syntax, [^/]* matches any string that doesn't contain a slash, \(/[^/]*\) matches a slash followed by any number of non-slash characters and makes the whole thing a group, and \1 matches the same text that is matched by the first group, therefore \(/[^/]*\)\1 matches two path segments with the same name. Add .* at the beginning to match the previous directory components.

With GNU find, FreeBSD find, macOS find or any other find that supports -regex:

find . -regex '.*\(/[^/]*\)\1'

Since there are generally fewer directories than files, let's look for all directories and then test whether they contain the required filename.

find . -type d -exec sh -c '
    for dirpath do
        filepath="$dirpath/${dirpath##*/}"
        [ -f "$filepath" ] && printf "%s\n" "$filepath"
    done' sh {} +

This would print the pathnames of all regular files (and symbolic links to regular files) that is located in a directory that has the same name as the file.

The test is done in a short in-line sh -c script that will get a number of directory pathnames as arguments. It iterates over each directory pathname and constructs a file pathname with the name that we're looking for. The ${dirpath##*/} in the code could be replaced by $(basename "$dirpath").

For the given example directory structure, this would output

./Example1/Example1
./Example2/Example2

To just test for any name, not just regular files, change the -f test to a -e test.

Use find -exec:

find . -type f \
  -exec sh -c '[ "$(basename "$1")" = "$(basename "$(dirname "$1")")" ]' find-sh {} \; \
  -print

The -print will only execute when the result of the [ ... ] inside the -exec sh -c '...' is true.

Output:

./Example1/Example1
./Example2/Example2