Krdytkyu

Define a category $mathcal{C}$ where the objects are finite sets and arrows are bijections between sets.

I want to define the functor "order" (Ord).

I have defined it as follows:

For a set $X in mathcal{C}$, $Ord(X)={sigma(x_1)<...<sigma(x_n) | sigma text{is a bijection}}$. ($sigma$ is a permutation).

For a bijection $f:Xlongrightarrow Y$, $Ord(f):Ord(X) longrightarrow Ord(Y): {sigma(x_1)<...<sigma(x_n) | sigma text{is a bijection}} mapsto {(fcirc sigma circ f^{-1})(x_1)<...<(fcirc sigma circ f^{-1})(x_n) | sigma text{is a bijection}}$.

I have defined before the functor $Perm: mathcal{C} longrightarrow Sets$.

For a set $X in mathcal{C}$, $Perm(X)={sigma: X longrightarrow X | text{$sigma$ is a bijection}}$.

For a bijection $f:Xlongrightarrow Y$, $Perm(f): Perm(X) longrightarrow Perm(Y): sigma mapsto fcirc sigma circ f^{-1}$.

And I have checked that $Perm$ is a functor. I have also checked that $Ord$ satisfies the conditions, but I do not know if its definition is a proper definition.

Any feedback?

From the comments, it is clear that you want $newcommandOrd{operatorname{Ord}}Ord$ to take a (finite) set $X$ (though there's no reason to restrict to finite sets) to the collection of all total orders on $X$.

To define $Ord$, it is probably best to just use words rather than notation. The notation you've used doesn't really make sense, at least it didn't make sense to me, nor I think to user 170039 (based on the comment).

Thus to define $Ord$ on objects, I would just write.

$Ord(X)$ is the collection of all total orders on $X$ (which as a collection of relations on $X$ is naturally identified with a subset of $mathcal{P}(Xtimes X)$). I would probably ordinarily omit the parenthetical, I've just included it to be clear about what the set theoretic definition of a total order is (i.e. that it is a special kind of relation).

Then to define $Ord(f):Ord(X)to Ord(Y)$ when $f:Xto Y$ is a bijection, I would say, let $<$ be a total order on $X$, then define $<_f$ by $y_1<_fy_2$ if and only if $f^{-1}(y_1)<f^{-1}(y_2)$. You can check that this definition makes $<_f$ a total order on $Y$, and that it is functorial.

Side note

If I were defining $Ord$, I would broaden the category of definition to all sets with morphisms all injections, in which case $Ord$ becomes a contravariant functor.

The definition of $Ord(X)$ remains the same, but now we define $Ord(iota)$, where $iota : Ahookrightarrow B$ is an injection as the map that takes a total order on $B$, $<$ and restricts it to $A$, so $<mapsto <_{|_A}$. Formally,
we define $a_1 <_{|_A} a_2$ if and only if $iota(a_1) < iota(a_2)$.

Again you can check that this is well defined and functorial.

Edit:

I suspect you are working on a problem, like the one here where you try to show that $operatorname{Perm}$ and $Ord$ are not naturally isomorphic. It's not very clear from your question, but it would explain why you're bringing up the functor $operatorname{Perm}$.

In which case, you should still use the definition of $Ord$ I gave above, but note that for finite sets with bijections, $Ord(X)$ is naturally isomorphic to $operatorname{Hom}(X,n)$, where $n$ is the finite ordinal with the same cardinality as $X$ (Note that it is important that the Homs are bijections for this to work.).