Permutations preserving a filtration property Suppose $\mathcal{A}$ is a non-empty family of sets of natural numbers size $n$ with the property that if $\{k_1<k_2< \dots <k_n\}\in \mathcal{A}$ and $j_i\leq k_i$ for all $1\leq i\leq n$, then $\{j_1,j_2,\dots, j_n\}\in \mathcal{A}$ as well. 
If $A=\{k_1, k_2, \dots,k_n\}\in\mathcal{A}$, and $f$ is a permutation of $\mathbb{N}$,  denote by $f(A)=\{f(k_1), f(k_2), \cdots, f(k_n)\}$, and by $f(\mathcal{A})=\{f(A): A\in\mathcal{A}\}$.  
If $f(\mathcal{A})$ has the same filtration property as above, does it follow that $\mathcal{A}=f(\mathcal{A})$?
This is trivial when $n=1$, but for me not so clear otherwise. 
 A: Yes, $f(A)=A$ in this case.
First let us show that the filtration property you describe implies a seemingly stronger property.

Claim: If $\{k_1,\dots, k_n\}\in A$ and $j_i$ are so that $j_i\leq k_i$ for all $i\leq n$ then $\{j_1,\dots, j_n\}\in A$.

Note that we do not assume that the $k_i$ or $j_i$ are increasing, though me may assume wlog that at least the $k_i$ are increasing by reordering. For $n=1$ this is clear, so suppose this is true for $n-1$. As we will show inductively that, after reodering the $j_i$ in increasing order as $j_{p(i)}$, we will have $j_{p(i)}\leq k_i$, we may as well suppose that for $i<n$, the $j_i$ are already increasing and $j_i\leq k_i$. If $j_{n-1}<j_n$, the assertion we want to prove is trivial, so me may say that there is $i<n$ with $j_i<j_n<j_{i+1}$. We now have that $j_n<j_{i+1}\leq k_{i+1}$. Furthermore for $i<l<n$, we get $j_{l}<j_{l+1}\leq k_{l+1}$. This shows that the reordered sequence of the $j$'s is pointwise bounded by the $k$'s and thus $\{j_1,\dots, j_n\}\in A$.
With this out of the way, let us prove $A\subseteq f(A)$ (of course assuming that $f(A)$ has the filtration property). Let $\{j_1, \dots, j_n\}\in A$. We will find $\{k_1,\dots, k_n\}\in f(A)$ (both not necessarily increasing) with $j_{p(i)}\leq k_i$ for all $i\leq n$ and some permutation $p$ of $\{1,\dots, n\}$. With the above claim, we get $\{j_1,\dots j_n\}\in f(A)$ as desired. 
By the pidgeonhole principle and as $f$ is injective, we can find for all $j_i$ distinct $l_i\leq j_i$ with $k_i:=f(l_i)\geq j_{p(i)}$ for some permutation $p$. By the above claim, $\{l_1,\dots, l_n\}\in A$ and thus $\{k_1,\dots, k_n\}\in f(A)$.
The reverse inclusion $f(A)\subseteq A$ now follows by applying the above to $A^\prime= f(A)$, and $f^\prime=f^{-1}$, which is okay as $f^\prime(A^\prime)= A$ has the filtration property, of course.
