Is it possible to do this in Awk?:

echo "eoe" | sed -nr '/^(.*)o\1$/p'

Not in standard awk (POSIX awk uses POSIX EREs which don't support back references, and \1 means the 0x1 character in awk, though there are some ambiguities). It's possible with busybox awk though using:

busybox awk '$0 ~ "^(.*)o\\1$"'

(what that may or may not do (whether that "\\1" should match a literal \1 or the 0x1 character or be unspecified) is unclear in the POSIX specification. In my reading it seems to imply it should match a 0x1 character, but it doesn't with /usr/xpg4/bin/sh on Solaris 11 for instance which is a certified OS (where it matches on a literal \1 instead))

With any awk, for that particular regexp, you could take another approach like:

awk 'length % 2 && \
       substr($0, (length+1)/2, 1) == "o" && \
       substr($0, 1, (length-1)/2) == substr($0, (length+3)/2)'

As mentioned above POSIX EREs don't support back-references. GNU sed with -r uses EREs, but that's GNU EREs that support back-references as an extension over the standard. What that means is that

grep -Ex '(.*)o\1'

(or same with egrep) is not portable. However:

grep -x '\(.*\)o\1'

is POSIX and portable. POSIX BREs do support back-references, as did historical implementations of grep. perl regexps or PCREs do support back references as well so you can do:

perl -lne 'print if /^(.*)o\1$/'

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