Consider the sets $X=\{1,2,4\}$ and $Y=\{A,B,C\}$ and then consider two functions $f:X\to X$ and $g:Y\to Y$ defined as $f=\{(1,4),(2,1),(4,1)\}$ and $g=\{(A,C),(B,A),(C,A)\}$. Certainly, these function have the same "structure", but what is it? What makes this functions more or less equal?

What I've noticed is that there exists a function $h:X\to Y$ such that $f=h^{-1}\circ g\circ h$. Must this hold for any two functions to have the same "structure"? What property does this make $f$ and $g$ have?

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    $\begingroup$ There are multiple ways of phrasing the equivalence. For example each map corresponds to a directed graph where an arrow goes from node x to node y if x maps to y. The directed graphs corresponding to f and g are isomorphic digraphs, with h the isomorphism. $\endgroup$ Apr 12, 2018 at 15:59
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    $\begingroup$ You cannot have the set $B$ be a member of itself. Please choose better names. $\endgroup$ Apr 12, 2018 at 16:25

2 Answers 2


Note that $h$ is a bijection. We say that $f$ and $g$ are conjugate maps. This notion is very useful in linear algebra and dynamical systems for instance, to reduce maps to simpler forms.

  • $\begingroup$ Maybe "conjugate self-maps" would be a better name. Since the defining property (formulated in the question) only works for maps whose domain and codomain are the same. $\endgroup$ Apr 12, 2018 at 16:29

You are correct - you can construct two composite functions $h \circ f:A \to B$ and $g \circ h:A \to B$ such that $(h \circ f)(x) = (g \circ h) (x) \quad \forall x \in A$. In other words $h \circ f = g \circ h$.

You could say that functions $f$ and $g$ are isomorphic. In category theory terms, $f, g$ and $h$ form a commutative diagram.

  • $\begingroup$ The term "isomorphic" should always specify a category if (like here) there is no obvious category one is working in. $\endgroup$ Apr 12, 2018 at 16:32
  • $\begingroup$ I know I'm late, but I think I found a mistake. The condition $h\circ f=g\circ h$ does not imply $h$ is an isomorphism, $h$ must have an inverse as lhf said. Otherwise, if $Im(h)=\{k\}$ and $g=\{(k,k)\}$, your condition states them as isomorphic but generally the structure is not preserved, as $f$ and $g$ may have different sizes. $\endgroup$
    – Garmekain
    Dec 11, 2019 at 15:54

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