Can these averaging formulas be expressed as transformations of the arithmetic mean? (Vaguely inspired by this question.)
Given three numbers $x,y,z$, their arithmetic mean is $m_a(x,y,z) = \frac{x+y+z}3$.
Their geometric mean is $m_g(x,y,z) = \sqrt[3]{xyz}$, but this is secretly the arithmetic mean in log-space: $m_g(x,y,z) = \log^{-1}(m_a(\log x,\log y,\log z))$.
We can also consider some "mixed" means:


*

*$m_{ag}(x,y,z) = \frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}3$.

*$m_{ga}(x,y,z) = \sqrt[3]{\left(\frac{x+y}2\right)\left(\frac{y+z}2\right)\left(\frac{z+x\vphantom{y}}2\right)}$.
I am curious whether these can also be expressed as the arithmetic mean under some transformation. That is, does there exist a function $f$ such that $m_{ag}(x,y,z) = f^{-1}(m_a(f(x),f(y),f(z)))$, and similarly for $m_{ga}$?
Update: Following the links in @Chappers' comment, it turns out that what I'm asking is whether the above formulas are quasi-arithmetic means a.k.a. generalized $f$-means.
 A: Münnich et al. (1999, 2000) give a complete characterization of quasi-arithmetic means, which Matkowski and Páles summarize as follows:

Let $n\geq2$ and let $M:I^n\to I$ [where $I\subseteq\mathbb R$ denotes a non-degenerated interval]. Then $M$ is an $n$-variable quasi-arithmetic mean ... for some continuous strictly monotone function $f:I\to\mathbb R$ if and only if
  
  
*
  
*$M$ is a continuous and symmetric function on $I^n$ 
  which is strictly increasing in each of its variables;
  
*$M$ is reflexive [that is, $M_n(x,\dots,x)=x$ for all $x\in I$];
  
*$M$ is bisymmetric, that is, for all $x_{i,j}\in I$ ($i,j\in\{1,\dots,n\}$), we have
  $$\begin{align}&M(M(x_{1,1},\dots,x_{1,n}),\dots,M(x_{n,1},\dots,x_{n,n}))\\
  {}={}&M(M(x_{1,1},\dots,x_{n,1}),\dots,M(x_{1,n},\dots,x_{n,n})).\end{align}$$
  

Taking $x_{i,j}$ to be given by the matrix $\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix}$, we find that neither $m_{ag}$ nor $m_{ga}$ are bisymmetric, so they cannot be quasi-arithmetic means.
