RSA by hand - did I do something wrong? (c = m on encryption)
In fact, your calculating is right. The problem is:

At first, $e=13=5\pmod 8$. 

Next, $d=5=e$ since $d\cdot e=1\pmod 8$. This means that the decryption key is identical to the encryption key. You should avoid this of course. 

Third, $7^{13}=7^5=(7^2)\cdot(7^2)\cdot 7\pmod{15}=(4\cdot 4)\cdot 7\pmod{15}=7\pmod{15}$. This is why your decryption outputs the same as the input. If you try another message, you may find that the cihertext is different to the plaintext.