When functions commute under composition
A classic result of Ritt shows that polynomials that commute under composition must be, up to a linear homeomorphism, either both powers of $x$, both iterates of the same polynomial, or both Chebychev polynomials. Actually Ritt proved a more general rational function case - follow the link. His work was motivated by work of Julia and Fatou's work on Julia sets of rational functions, e.g. see here for a modern presentation.
According to Wikipedia, a set of diagonalizable matrices commute if and only if they are simultaneously diagonalizable. There is a far-reaching generalization, namely the Gelfand representation theorem.
The Gelfand representation theorem for commutative $C^*$ algebras represents every commutative $C^*$ algebra as an algebra of functions with pointwise multiplication; the domain of the latter algebra is the spectrum of the former algebra.
This question may also be related to how certain functions behave under functions of their variables. In this context, the property of commuting with binary operators, such as addition and multiplication, can be used to define classes of functions:
additive commutation: if $g(x, y) = x + y$, then $f\big(g(x, y)\big) = g\big(f(x),\ f(y)\big)$ if and only if $f(x + y) = f(x) + f(y)$ thus $f$ is a homogeneous linear function of the form $f(x; a) \equiv ax$
multiplicative commutation: if $g(x, y) = xy$, then $f\big( g(x, y) \big) = g\big(f(x),\ f(y)\big)$ if and only if $f(xy) = f(x)f(y)$ thus $f$ is "scale invariant" i.e. a power law of the form $f(x; a) \equiv x^a$
log-additive commutation: if $g(x, y) = x + y$, then $\log f\big( g(x, y) \big) = g\big( \log f(x),\ \log f(y) \big)$ if and only if $f(x + y) = f(x)f(y)$ thus $f$ is an exponential function of the form $f(x; a) \equiv \exp(ax)$
The last item (3) involves a third function (the logarithm) which when denoted as $h$ gives
$h\big(f[g(x, y)]\big) = g\big(h[f(x)],\ h[f(y)]\big)$
or
$h \circ f \circ g(x, y) = g\big(h \circ f(x),\ h \circ f(y)\big).$
Since $h \circ f$ occurs on both sides, we can denote this as $\tilde f$ to get
$\tilde f \big( g(x, y) \big) = g \big( \tilde f(x), \tilde f(y) \big)$
which has the same form as item (1) above. From this perspective, items (1) and (3) above can be seen as being isomorphic under the $\exp$ and $\log$ pair of invertible mappings.