Obviously cannot check all of them

There are quite a few cases for this.

If $n$ ends in 1, 0, 5, 6 or 9 then $n^n$ always ends in the same ones digit as $n$ . Since there are 5 such occurrences of such numbers from 1 to 10, then

no. of occurrences from 1-2010 = $5(\frac {2010-10} {10} +1)=1005$

Therefore, no. of occurrences of such from 1-2015 is 1007 (2011 and 2015 included.)

If $n$ ends in 3 or 7, then $n^n$ ends in the same ones digit as $n$ if and only if $n \equiv 13 \pmod {20}$ or $n \equiv 17 \pmod {20}$ . Since there are 2 occurrences of such numbers from 1-20, therefore

no. of occurrences from 1-2000 = $2(\frac {2000-20} {20} +1)=200$

Therefore, no. of occurrences of such from 1-2015 is 201 (2013 included.)

If $n$ ends in either 2, 4 or 8, then $n^n$ will never end in the same digit as $n$ . As such, there are 0 occurrences of $n$ which ends in any of there numbers.

Therefore, our answer is $1007 + 201 = 1208$