Find the index of reflection of a substring of a palindrome Problem Statement:
You will receive a substring of a palindromic string. You must return the index of the substring which marks the point of reflection of the original string. You are only provided the substring, which is not necessarily a palindrome because it is not necessarily centered about the middle of the original palindromic string.
Input: Substring of length 2n + 1,  1 &lt;= n &lt;= 1000, which encompasses the center of reflection of some larger palindromic string.
Output: Return the index of the substring which marks the point of reflection. May be 0-indexed or 1-indexed.
Test cases (character at desired index is bold to show the desired output more clearly):
Input

manaplanacanalpan
caroracati
wlatemymeta
nasanita

Output

9 (the full string was “a man a plan a canal panama”, with spaces removed)
4 (the full string was “was it a car or a cat i saw”, with spaces removed)
6 (the full string was “mr owl are my metal worm”, with spaces removed)
2 (the full string was “oozy rat in a sanitary zoo”, with spaces removed)

Winning Criterion: This is code-golf Shortest code in bytes wins.
Assumption: Assume there is just a single candidate palindrome
Restriction:0 and 2n (last index) are invalid output.
J, 26 25 bytes
[:(-:@i.&gt;./)[:(*/*#)/.=/~

Try it online!
how
=/~ creates a function table comparing all possible pairs of characters.  Taking 'nasanita' as an example:
1 0 0 0 1 0 0 0
0 1 0 1 0 0 0 1
0 0 1 0 0 0 0 0
0 1 0 1 0 0 0 1
1 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 1 0 1 0 0 0 1

Notice the /-direction diagonals.  If one of those is all ones, it corresponds to an embedded palindrome.
J has an adverb /. which lets us to apply a verb to each diagonal.
We choose the verb (*/*#) -- the sum multiplied by the length.  Thus it will be the length of the diagonal if it's all ones, and 0 otherwise:
1 0 0 0 5 0 0 0 0 0 0 0 0 0 1

i.&gt;./ Find the index of the max element within that list.  In this case the index of 5 is 4.
-:@ And divide it by 2.