Are all well behaved "mean" functions on $\mathbb{R}^+$ equivalent? Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties:


*

*$M(a,a)=a\qquad$ (identity)

*$M(a,b)=M(b,a)\qquad$ (commutativity).
and possibly


*$M(M(a,b),M(a,c))=M(a,M(b,c))\qquad$ (weak associativity)

*$M(M(a,b),M(c,d))=M(M(a,c),M(b,d))\qquad$ (strong associativity)

*$a\ne b \implies a\ne M(a,b)\ne b\qquad$ (sharpness).
When $S$ is an abelian groupoid or an ordered set or a topological space, $M$ can have additional specific requirements, such as:


*$M(ac,bc)=cM(a,b)\qquad$  (homogeneousness - see UPDATE)

*$a < b \implies a\le M(a,b) \le b\qquad$ (order preservation - see UPDATE)

*continuity.

This counteraxample is wrong. After fixing my gawk code I found that there (3) always implies (4) in a set of 5. Thanks to Eric Wofsey for noticing.
In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$:
$$
\begin{array}{c|ccccc}
M & a & b & c & d & e\\
\hline
a & a & a & a & a & a\\
b & a & b & d & c & a\\
c & a & d & c & b & a\\
d & a & c & b & d & e\\
e & a & a & a & e & e\\
\end{array}$$
where $M(M(b,c),M(d,e))  \ne M(M(b,d),M(c,e))$.

Here are some of the questions that come to mind.
Q1.  Is there a finite example where (3) and (5) hold, but not (4)? I know that $S$ will need to have at least 6 elements.
Q2. Does $M$ in the above example naturally extend to a mean in $\mathbb{R}[a,b,c,d,e]$ where both (3) and (6) hold?
Another example: if $A$ and $G$ are the arithmetic and geometric means on $\mathbb{R}^+$, it's easy to check that the mean function $M(x,y)=G(A(x,y),G(x,y))$ satisfies all the properties except (3) and (4).
Q3. Assuming all of the above properties except (4) hold for $M$ on $\mathbb{R}^+$, does (4) follow?
Q4. My starting point leading to this post: if all the above properties, including (4), hold for $M$ on $\mathbb{R}^+$, does it follow that $M$ is equivalent to the arithmetic mean, in the sense that $M(x,y)=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ for some continous strictly monotonic function $f: \mathbb{R}^+ \to \mathbb{R}$?
I welcome suggestions for improvements to this post and references to relevant work. 
UPDATE. It turns out that neither (6) nor (7) are needed anywhere, at least for the questions explored in this thread.

This is superseded by Eric Wofsey's answer
UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{y_1+y_2}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$.

 A: Questions like this have a long history, going back to 1930s work of Kolmogorov, Nagumo, and Hardy, Littlewood and Pólya. There's a modern account (plus some new stuff) in Chapter 5 of my book Entropy and Diversity: The Axiomatic Approach. The introduction to the chapter very briefly summarizes a little of the history. @EricWofsey's nice answer rediscovers some of the techniques and results used by those old masters (though I don't mean to suggest it's only rediscovery).
More specifically, Theorems 5.3.2, 5.3.3, 5.4.7 and 5.4.9 of my book give four different sets of conditions on a "mean" that turn out to be equivalent to it being a power mean of some order.
A: Define a mean algebra to be a set $S$ with an binary operation $M$ satisfying (1), (2), and (4).  We can define $M(a,b,c,d)=M(M(a,b),M(c,d))$ and this will depend only on the multiset $\{a,b,c,d\}$.  More generally, we can think of $M$ as an operation defined on multisets of size $2^n$ for any $n>0$ (and this is well-defined by an easy induction on $n$ using (2) and (4)).  The free mean algebra on two generators can be identified with the dyadic rationals between $0$ and $1$ with the arithmetic mean (the generators being $0$ and $1$); freeness of this algebra is easy to see when you write $k/2^n$ as $M(0,0,\dots,0,1,1,\dots,1)$ (with $k$ $0$s and $2^n-k$ $1$s).  Denote this algebra by $Q$.  Write $I$ for the algebra $[0,1]$ with the arithmetic mean; this can be thought of as a sort of completion of $Q$.
Suppose $A$ is a mean algebra with underlying set $[0,1]$ that further satisfies (5), (7), and (8).  There is a unique mean-preserving map $f_0:Q\to A$ satisfying $f_0(0)=0$ and $f_0(1)=1$, and it follows easily from (5) and (7) that it is injective and order-preserving.  The inverse of $f_0$ extends uniquely to an order-preserving surjection $g:A\to I$ (every element of $A$ defines a Dedekind cut in $Q$ via $f_0$).  By (7) and (8) $g$ will also be mean-preserving, and so (5) implies $g$ is injective.  We thus conclude that $A$ is isomorphic to $I$ as an ordered mean algebra, and furthermore this isomorphism is unique.
Now suppose $A$ is a mean algebra with underlying set $\mathbb{R}$ satisfying (5), (7), and (8).  Every compact subinterval of $A$ these has a unique order-preserving isomorphism to $I$, and by uniqueness we can glue these isomorphisms together to get an order-preserving isomorphism between $A$ and an open subinterval of $\mathbb{R}$ with the arithmetic mean.  Note that there are actually 4 distinct isomorphism classes of such structures (with respect to both the order and the mean): $(-\infty,\infty)$, $(-\infty, 0)$, $(0,\infty)$, and $(0,1)$.  In particular, this gives an affirmative answer to Q4 (without assuming (6)).  Furthermore, the isomorphism in question is clearly unique up to composition with affine maps $\mathbb{R}\to\mathbb{R}$.
As for your desire to only assume (3) and not (4), the only place where I used (4) is in asserting that $Q$ is free.  Let $F$ be the free algebra on $\{0,1\}$ assuming (3) instead of (4); we wish to prove that the canonical map $F\to Q$ is an isomorphism.  Every element of $F$ can be represented as a full binary tree of some height $n$ where each of the $2^n$ leaves is labelled by $0$ or $1$, and at each juncture of the tree we apply $M$.  It suffices to show that if two such trees have the same number of leaves that are 1, then they represent equal elements of $F$.  We prove this by induction on $n$; the cases $n\leq 1$ are trivial.
Let $x\in F$ be represented by such a tree of height $n>1$ that has $i$ leaves that are $1$.  WLOG $i\leq 2^{n-1}$ (otherwise we can just swap the roles of $0$ and $1$).  Let $y$ be represented by a tree that looks the same as $x$'s, except that all the $0$s are on the left and all the $1$s are on the right.  For instance, if $n=i=3$ and $$x=M(M(M(1,0),M(0,1)),M(M(0,1),M(0,0))),$$ then $$y=M(M(M(0,0),M(0,0)),M(M(0,1),M(1,1))).$$
It suffices to prove that $x=y$, since $y$ depends only on $n$ and $i$.  Furthermore, by induction, it suffices to prove that $y=M(z,w)$ where $z$ and $w$ each have as many $1$s as the left and right halves of $x$, respectively (since $z$ and $w$ can then be transformed to look the same as the two halves of $x$).  Let $j$ be the number of $1$s in the left half of $x$; WLOG $j\leq 2^{n-2}$ (if not, switch the two halves of $x$).  By the induction hypothesis, we can write the right half of $y$ as $M(b,c)$, where $b$ has $j$ $1s$.  Also, the left half of $y$ is $M(a,a)$, where $a$ is the tree of height $n-2$ consisting entirely of $0$s.  We thus have $y=M(a,M(b,c))$, and so by (3) we can rewrite it as $y=M(M(a,b),M(a,c))$.  Setting $z=M(a,b)$ and $w=M(a,c)$, the proof is complete.
As a final remark, I'm not sure what happens when you additionally assume (6).  Certainly most functions $f$ as in your Q4 will not give rise to a mean satisfying (6); the only ones I know of that do are $f(x)=x^p$ for $p\neq0$ and $f(x)=\log x$, up to composition with affine maps.  Note that $f(x)=\log x$ (corresponding to the geometric mean) can be thought of as the $p=0$ case; indeed, the mean obtained from it is equal to the limit of the $x^p$ means as $p\to 0$, and these functions $f$ (including their compositions with affine maps) are exactly the solutions of the differential equation $(xf''/f')'=0$.  Perhaps if you assume $f$ is sufficiently differentiable you could prove it must be of this form by differentiating the functional equation you get from associativity.
EDIT: Here's a proof that if (6) holds, then $f(x)$ must be of the form $ax^p+b$ or $a\log x+b$ and hence $M$ is either $((x^p+y^p)/2)^{1/p}$ or $\sqrt{xy}$.  Suppose that (6) holds and $f:\mathbb{R}_+\to U\subseteq\mathbb{R}$ is an isomorphism from $M$ to the arithmetic mean on some open interval $U$.  For $c>0$, consider the map $x\mapsto f(cf^{-1}(x))$.  This is a mean-preserving automorphism of $U$, so it must be of the form $x\mapsto A(c)x+B(c)$ for some constants $A(c)$ and $B(c)$ depending on $c$.  That is, we have $f(cx)=A(c)f(x)+B(c)$.
Write $r=A(e)$; then it is easy to show that if $c=e^q$ for $q\in\mathbb{Q}$, we have $A(c)=r^q$.  By continuity, it follows that $A(c)=r^{\log c}=c^p$ for all $c$, where $p=\log r$.  Suppose that $p=0$, so $f(cx)=f(x)+B(c)$.  Then $B(cd)=B(c)+B(d)$ and it follows by continuity that $B(c)=a\log c$ for some $a$.  Setting $b=f(1)$, we then get $f(c)=a\log c +b$, as desired.
Now suppose that $p\neq 0$.  Let $s=B(e)$; then by induction we have $$B(e^n)=s\frac{r^{n+1}-1}{r-1}$$ for $n\in \mathbb{N}$.  But by the same argument using $e^{1/m}$ instead of $e$ for some integer $m>0$, we must have $$B(e^n)=B(e^{1/m})\frac{r^{n+1}-1}{r^{1/m}-1}.$$  It follows that the first formula is also valid when $n=1/m$, and it is now easy to show that it is valid for all rational $n$.  By continuity, it holds for all real $n$, and we can rewrite it to get $$B(c)=s\frac{e^pc^p-1}{r-1}=tc^p+b$$ for some constants $t$ and $b$.  Thus we have $f(cx)=c^pf(x)+tc^p+b$, and setting $x=1$ and $a=f(1)+t$ we get $f(c)=ac^p+b$, as desired.
