RSA by hand - did I do something wrong? (c = m on encryption) to understand RSA better I am doing a little calculation by hand, this is what I got:

Choosing:

$p = 3\\
q = 5\\
n = 15\\
\varphi(p\cdot q) = 2 \cdot 4 = 8\\
e &gt; \varphi(n) \implies e = 13\\
e \cdot d = 1 \pmod 8\\
13 \cdot d = 1 \pmod 8\\
(13 \cdot 5) \mod 8 = 1$

So there has to be $m \le n$ and I choose $m = 7$

Encryption:

$c = m^e \mod n\\
c = 7^{13} \mod 15\\
c = 7$

And if I decrypt it's nice coming back to 7. But it looks a bit weird to me having $m = c$ but I don't find what I should be doing wrong.. ?
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.