Notation for set of all possible bijections I have two finite sets: $S$and $S'$. I know $\mathrm{card}(S) = \mathrm{card}(S')$
I have a function $colour()$ that produces some colour for each element of $S$and $S'$.
I know that there is at least one "complete" bijection from $S$ to $S'$ that preserves colour, but there are as many as $\mathrm{card}(S)!$
I need some succinct notation to enumerate all such bijections that:


*

*are from $S$ to $S'$

*preserve colour

*cover all elements of $S$ and $S'$


The more succinct and intuitive and widely understandable the notation, the more preferable (I'm aiming for one line in an algorithm).
What I'm current looking at is something like this:
$M := \{ \mu : S \xrightarrow{1:1} S' \mid s \mapsto s' : colour(s) = colour(s') \}$
... but I'm not sure this is "standard" / well-understood / complete on $S,S'$ / etc., so I'm looking for alternatives or feedback.

EDIT I really need to construct the set on one line without prose and I really don't think it needs to be so complicated as some (not all) of the answers suggest. :) 
For the moment, very much inspired by Brian's answer, I'm going with simply:
$M := \{ \mu : S \xrightarrow{1:1} S' \mid colour(\mu(s)) = colour(s)\text{ for all }s\in S\}$
I hope this is clear and intuitive: $M$ is the set of all 1:1 mappings $\mu$ between $S$ and $S'$ where $\mu$ preserves colour.
 A: I’d prefer to do it in two stages, and with a few more words:

Let $M^+$ be the set of bijections from $S$ to $S'$, and let $$M=\left\{\mu\in M^+:\operatorname{colour}(s)=\operatorname{colour}\big(\mu(s)\big)\text{ for all }s\in S\right\}\;.$$

A: Let $S_c = \{s \in S: colour(s) = c\}$.
We can express the number of such bijections as $\vert M \vert = \prod_c{(\vert S_c \vert !)}$.
The same notation can be abused to say $M = F(\prod_c{S_c!})$ where $F:S \rightarrow S'$ is the color-preserving bijection we are guaranteed to have.  Or if $S = S'$, simply $M = \prod_c{( S_c !)}$
EDIT: One way to make it less ambiguous, is to first define the factorial of a set as the set of all bijections between $S$ and itself.  Now for the $S = S'$ case we can write instead:
$M = \bigcup_c{(S_c!)}$.
And there is no longer any ambiguity, since the union of two functions with disjoint domains is a function defined on the union of those domains specializing to whichever corresponding function.  Now if it also understood that a function composed with a set of functions denotes the set comprised of that function composed with each element of that set, then for the general case $S \ne S'$ we can write without ambiguity:
$M = F(\bigcup_c{(S_c!)})$
A: If $C$ is the set of colors, then we can introduce the notion of a $C$-colored set. A $C$-colored set is hereby defined to be a pair $(S,c)$ where $S$ is a set and $c:S \to C$ (this is the function you called color).
A homomorphism of colored sets from $(S,c)$ to $(S',c')$ is a function $f:S \to S'$ with the additional property that $c(x) = c'(f(x))$.
An isomorphism is an invertible homomorphism. If $f$ is a homomorphism of $C$-colored sets, then it is a bijection of sets if and only if it is an isomorphism of $C$-colored sets.
If $X$ and $Y$ are colored sets -- that is, we have $X = (X', c)$ and $Y = (Y',d)$ where $c:X' \to C$ and $d : Y' \to C$, then the set of all isomorphisms of colored sets would be called
$$ \text{Iso}(X, Y) $$
