(a)+(b): note that we get that $2f(f(x))=f(f(f(x)))\,\diamondsuit$, since $H \to H$, we get that for some $z\in H$, we have that $f(z)=2z$. Setting $x=z$ in (a), we have that $f(2z)=z$. Thus, $f(f(z))=f(2z)=z$ and $f(f(2z))=f(z)=2z$. Setting $x=2z$ in $\diamondsuit$, we get that $f(2z)=4z=z\implies z=0$. Hence, $f(f(x))=0$ for all $x$, otherwise if $f(f(x))=a\neq 0$ for some $x$, then we have contradiction since $f(z)=2z\implies z=0$. Now by the first one we have that $f(x)=x$, but since $f(f(x))=f(x)=x=0$ for all $x$, we have a contradiction. No such $f$.

@above. (a),(b),(c) are separated.

(b)+(c): note that we get that $f(f(x))=2f(f(f(x)))\,\diamondsuit$, since $H \to H$, we get that for some $z\in H$, we have that $f(z)=\frac12z$. Setting $x=z$ in (b) and (c), we have that $f(f(0.5z))=f(0.5z)-0.5z$ and $3f(f(0.5z))=2f(0.5z)-0.5z$, thus $3(f(0.5z)-0.5z)=2f(0.5z)-0.5z\implies f(0.5z)=z$. Thus, $f(f(0.5z))=0.5z$. Setting $x=0.5z$ in $\diamondsuit$, we get that $2f(0.5z)=2z=0.5z\implies z=0$. Hence, $f(f(x))=0$ for all $x$, otherwise if $f(f(x))=a\neq 0$ for some $x$, then we have contradiction since $f(z)=0.5z\implies z=0$. By (c), setting $x=0$, we get that $-2f(0)=0\implies f(0)=0$ Now by (b) we have that $f(x)=x+f(0)=x$, but since $f(f(x))=f(x)=x=0$ for all $x$, we have a contradiction. No such $f$.

Rickyminer wrote: @above. (a),(b),(c) are separated. Oh really, what ahahha

those are 3 different questions, parts a and b were posted for the category E, parts b and c were posted for the category E+