Problem: \def\di{\mathbin{\diamondsuit}}
Find all binary operations $\di \colon \RR_{>0} \times \RR_{>0} \to \RR_{>0}$
(meaning $\di$ takes pairs of
positive real numbers to positive real numbers)
such that for any real numbers $a,b,c > 0$,
\begin{itemize}
  \ii the equation $a \di (b \di c) = (a \di b) \cdot c$ holds; and
  \ii if $a \ge 1$ then $a \di a \ge 1$.
\end{itemize}

Solution: \def\di{\mathbin{\diamondsuit}}
The answer is only multiplication and division,
which both obviously work.

We present two approaches,
one appealing to theorems on Cauchy's functional equation,
and one which avoids it.

\paragraph{First solution using Cauchy FE.}
We prove:
\begin{claim*}
  We have $a \di b = a f(b)$ where $f$ is
  some involutive and totally multiplicative function.
  (In fact, this classifies all functions
  satisfying the first condition completely.)
\end{claim*}
\begin{proof}
  Let $P(a,b,c)$ denote the
  assertion $a \di (b \di c) = (a \di b) \cdot c$.
  \begin{itemize}
    \ii Note that for any $x$,
    the function $y \mapsto x \di y$ is injective,
    because if $x \di y_1 = x \di y_2$ then take $P(1, x, y_i)$
    to get $y_1 = y_2$.

    \ii Take $P(1,x,1)$ and injectivity to get $x \di 1 = x$.
    \ii Take $P(1,1,y)$ to get $1 \di (1 \di y) = y$.
    \ii Take $P(x, 1, 1 \di y)$ to get
    \[ x \di y = x \cdot (1 \di y). \]
  \end{itemize}
  Henceforth let us define $f(y) = 1 \di y$, so
  $f(1) = 1$, $f$ is involutive and
  \[ x \di y = xf(y). \]

  Plugging this into the original condition now gives
  $f(bf(c)) = f(b)c$,
  which (since $f$ is an involution)
  gives $f$ completely multiplicative.
\end{proof}

In particular, $f(1) = 1$.
We are now interested only in the second condition,
which reads $f(x) \ge 1/x$ for $x \ge 1$.

Define the function
\[ g(t) = \log f(e^t) \]
so that $g$ is additive,
and also $g(t) \ge -t$ for all $t \ge 0$.
We appeal to the following theorem:
\begin{lemma*}
  If $h \colon \RR \to \RR$ is an additive function
  which is not linear,
  then it is \emph{dense} in the plane:
  for any point $(x_0, y_0)$ and $\eps > 0$
  there exists $(x,y)$ such that $h(x) = y$
  and $\sqrt{(x-x_0)^2 + (y-y_0)^2} < \eps$.
\end{lemma*}
Applying this lemma with the fact
that $g(t) \ge -t$ implies readily that $g$ is linear.
In other words, $f$ is of the form $f(x) = x^r$
for some fixed real number $r$.
It is easy to check $r =\pm 1$ which finishes.

\paragraph{Second solution manually.}
As before we arrive at $a \di b = af(b)$,
with $f$ an involutive and totally multiplicative function.

We prove that:
\begin{claim*}
  For any $a > 0$, we have $f(a) \in \{1/a, a\}$.
\end{claim*}
\begin{proof}
  WLOG $b > 1$, and suppose $f(b) = a \ge 1/b$ hence $f(a) = b$.

  Assume that $ab > 1$; we show $a = b$.
  Note that for integers $m$ and $n$ with $a^n b^m \ge 1$, we must have
  \[ a^m b^n = f(b)^m f(a)^n
    = f(a^n b^m) \ge \frac{1}{a^n b^m} \implies (ab)^{m+n} \ge 1 \]
  and thus we have arrived at the proposition
  \[ m+n < 0 \implies n \log_b a + m < 0 \]
  for all integers $m$ and $n$.
  Due to the density of $\QQ$ in the real numbers,
  this can only happen if $\log_b a = 1$ or $a = b$.
\end{proof}

\begin{claim*}
  The function $f$ is continuous.
\end{claim*}
\begin{proof}
  Indeed, it's equivalent to show $g(t) = \log f(e^t)$ is continuous, and we have that
  \[
    \left\lvert g(t)-g(s) \right\rvert = \left\lvert \log f(e^{t-s}) \right\rvert
    = \left\lvert t-s \right\rvert
  \]
  since $f(e^{t-s}) = e^{\pm |t-s|}$.
  Therefore $g$ is Lipschitz.
  Hence $g$ continuous, and $f$ is too.
\end{proof}

Finally, we have from $f$ multiplicative that
\[ f(2^q) = f(2)^q \]
for every rational number $q$, say.
As $f$ is continuous this implies $f(x) \equiv x$ or $f(x) \equiv 1/x$ identically
(depending on whether $f(2) = 2$ or $f(2)=1/2$, respectively).

Therefore, $a \di b = ab$ or $a \di b = a \div b$, as needed.

\begin{remark*}
  The Lipschitz condition is one of several other ways to proceed.
  The point is that if $f(2) = 2$ (say),
  and $x/2^q$ is close to $1$,
  then $f(x)/2^q = f(x/2^q)$ is close to $1$,
  which is enough to force $f(x) = x$ rather than $f(x) = 1/x$.
\end{remark*}

\begin{remark*}
  Compare to AMC 10A 2016 \#23,
  where the second condition is $a \di a = 1$.
\end{remark*}