Let $f_m$: $\mathbb{R}_{\geq 0}^m \to \mathbb{R}_{\geq 0}$ be a series of functions that satisfy symmetry (when permuting indices), strong monotonicity (in every entry), homogeneity of degree 1,
$f_2(0,1)=\frac{1}{2}$ and for all $k<m$: $f_m(x)=f_m(f_k(x_1,...,x_k),...,f_k(x_1,...,x_k),x_{k+1},...,x_m)$
Show that for all $m$, $\;\; f_m(x)=\frac{x_1+...x_m}{m}$.
My attempt:
By homogeneity $f_2(x,0)=f_2(0,x)=xf_2(0,1)=\frac{x}{2}$. I noticed $f_1=$id, since $f_2(x,0)=f_2(f_1(x),0)$ implies $f_1(x)=x$: Indeed, injectivity in every entry follows from strong monotonicity: if by contrary $f_1(x)>x$ then $f_2(f_1(x),0)>f_2(x,0)$, and similarly we can't have $f_1(x)<x$, so $f_1(x)=x$ for all x.
I couldn't get any further,though: it seems impossible to use it to conclude anything decisive about the higher m's.
Without the "boundary condition", the set of solutions is closed under addition. But it isn't closed under subtraction , so it doesn't help.
We calculate $C_m=f_m(1,1,1, \dots)$:
$C_{2m}=f_{2m}(1,1,\dots)=f_{2m}(f_m(1,1,\dots,),\dots)= C_m C_{2m}$
The monotonicity ensures that $C_{2m}\not=0$, so we get $C_m=1$ for all $m$, homogeneity immediately implies $f_m(x,x,\dots,x)=x$.
$$d_{ab}=f_{ab}(1,0,\dots,0)=f_{ab}(f_a(1,0,\dots,0),\dots,f_a(1,0,\dots,0),0,\dots,0)=d_a f_{ab}(f_b(1,0,\dots,0),\dots,f_b(1,0,\dots,0))=d_ad_bf_{ab}(1,\dots,1)=d_ad_b$$
$$d_m=f_m(1,0,\dots,0)=f_m(f_{m-1}(1,0,\dots,0),\dots,f_{m-1}(1,0,\dots,0),0)=d_{m-1}f_m(1,\dots,1,0)<d_{m-1}f_m(1,\dots,1)=d_{m-1}.$$
For given integers $m$ and $k$, choose $l$ so that $$2^l\le m^k < 2^{l+1},$$ that is $l=\lfloor k \log_2(m)\rfloor$.
We know that $$d_{2^l} \ge d_{m^k} > d_{2^{l+1}}$$ which implies $$ 2^l \le \frac 1 {d_m^k} < 2^{l+1}.$$
Therefore, $l=\lfloor k \log_2(\frac 1{d_m})\rfloor$ and we get for all $k \in \mathbb N$: $$\lfloor k \log_2(m)\rfloor = \lfloor k \log_2(\frac 1{d_m})\rfloor$$
This implies $m=\frac 1{d_m}$ and we are done. (Just choose powers of ten for $k$ to see that the two logarithms share arbitrarily many decimals.)
Let $a,b$ be two positive integers. $$\frac1{a+b}2f_2(a,b)=d_{a+b}2f_2(a,b)= 2f_2(a,b)f_{a+b}(1,0,\dots,0)=2f_2(a,b)f_{a+b}(\frac12,\frac12,0,\dots,0)= f_{a+b}(a,b,0,0,\dots)=f_{a+b}(f_a(a,0,0,\dots),\dots,f_b(b,0,0,\dots))=f_{a+b}(1,1,\dots)=1.$$
Therefore, $f_2(a,b)=\frac{a+b}2$ for any positive integers $a,b$. By homogeneity, $f_2(1,z)=\frac{1+z}2$ for any positive rational $z$.
Monotonicity immediately implies that the identity is true for any positive real $z$. Homogeneity gives the desired result for $f_2(x,y)$.
$$f_m(x_1,x_2,\dots,x_m)= f_m(x_1,x_2,\dots,x_{m-2},\frac{x_{m-1}+x_m}2,\frac{x_{m-1}+x_m}2)=f_m(x_1,x_2,\dots,x_{m-2},x_{m-1}+x_m,0)=f_m(x_1,x_2,\dots,x_{m-2}+x_{m-1}+x_m,0,0)=\dots=f_m(x_1+x_2+\dots+x_m,0,\dots,0)=\dfrac{x_1+x_2+\dots +x_m}m.$$
The same proof should work for other means like the quadratic mean if we choose the normalization of $f_2$ to correspond to it.