Here are two inequivalent binary codes with the same distance enumerator. Code A consists of the codewords $a=00000000000$, $b=11110000000$, $c=11111111100$, $d=11000110011$; code R consists of the codewords $r=0000000000$, $s=1111000000$, $t=0111111111$, $u=1000111100$. Writing $xy$ for the Hamming distance between words $x$ and $y$, we have $ab=rs=4$, $bc=ru=5$, $cd=tu=6$, $ac=rt=9$, and $ad=bd=st=su=7$, so the distance enumerators are identical. But code A has the 7-7-4 triangle $adb$, while code R has the 7-7-6 triangle $tsu$, so there is no distance-preserving bijection between the two codes.
I'm sure there are shorter examples.
EDIT: a much shorter example: Code A consists of the codewords $a=000$, $b=100$, $c=011$, $d=111$; code R consists of the codewords $r=0000$, $s=1000$, $t=0100$, $u=0011$. Both codes have two pairs of words at each of the distances 1, 2, and 3; R has a 1-1-2 triangle, A doesn't.
MORE EDIT: In response to Jyrki's objections, here's one where both codes are of the same length, and no component is constant over all code words. $a=0000$, $b=1000$, $c=1110$, $d=1111$; $r=0000$, $s=0111$, $t=1110$, $u=1111$. Each code has two pairs at distance 1 and two at distance 3, and one each at distances 2 and 4; first code has a 1-3-4 triangle, $abd$, second code doesn't.