It is a slightly long, but it is a single command-line. This looks at the contents of the files and compares them using a cryptographic hash (md5sum).
find . -type f -exec md5sum {} + | sort | sed 's/ */!/1' | awk -F\| 'BEGIN{first=1}{if($1==lastid){if(first){first=0;print lastid, lastfile}print$1, $2} else first=1; lastid=$1;lastfile=$2}'
As I said, this is a little long...
The find runs md5sum against all files in the current directory tree. Then the output is sortd by the md5 hash. Since whitespace could be in the filenames, the sed changes the first field separator (two spaces) to a vertical pipe (very unlikely to be in a filename).
The last awk command tracks three variables: lastid = the md5 hash from the previous entry, lastfile = the filename from previous entry, and first = lastid was first time seen.
The output includes the hash so you can see which files are duplicates of each other.
This does not indicate if files are hard links (same inode, different name); it will just compare the contents.
Update: corrected based on just basename of file.
find . -type f -print | sed 's!.*/\(.*\)\.[^.]*$!\1|&!' | awk -F\| '{i=indices[$1]++;found[$1,i]=$2}END{for(bname in indices){if(indices[bname]>1){for(i=0;i<indices[bname];i++){print found[bname,i]}}}}'
Here, the find just lists the filenames, the sed takes the basename component of the pathname and creates a two field table with the basename and the full pathname. The awk then creates a table ("found") of the pathnames seen, indexed by the basename and the item number; the "indices" array keeps track of how many of that basename have been seen. The "END" clause then prints out any duplicate basenames found.
