defining functor $Ord$ 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 $f:X\longrightarrow Y$, $Ord(f):Ord(X) \longrightarrow Ord(Y): \{\sigma(x_1)<...<\sigma(x_n) | \sigma \text{is a bijection}\} \mapsto \{(f\circ \sigma \circ f^{-1})(x_1)<...<(f\circ \sigma \circ f^{-1})(x_n) | \sigma \text{is a bijection}\}$.
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 $f:X\longrightarrow Y$, $Perm(f): Perm(X) \longrightarrow Perm(Y): \sigma \mapsto f\circ \sigma \circ f^{-1}$.
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?
 A: From the comments, it is clear that you want $\newcommand\Ord{\operatorname{Ord}}\Ord$ 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 $\mathcal{P}(X\times X)$). 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 $f:X\to Y$ 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 $\iota : A\hookrightarrow B$ 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.).
