Okay I was thinking it over and you are right I don't think I made any errors. I don't know why I used n/2. I think I was just in a hurry to write it out lol
Suppose $\displaystyle n $ is even and $\displaystyle n=2^am $ where $\displaystyle m $ is odd.
$\displaystyle \phi(2n)=\phi(2^{a+1}m)=\phi(2^{a+1})\phi(m) = 2^a\phi(m) $
$\displaystyle 2\phi(n)=2\phi(2^am)=2\phi(2^a)\phi(m)=2\cdot2^{a-1}\phi(m)=2^a\phi(m) $
Thus $\displaystyle \phi(2n)=2^a\phi(m) $ and $\displaystyle 2\phi(n)=2^a\phi(m)\implies \phi(2n)=2\phi(n) $