Filters invariant under bijections Which filters on a set $A$ are invariant under every bijection $A\leftrightarrow A$?
I found the following examples of invariant filters:


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*$\{ A\setminus K \mid K\in\mathscr{P}A, \operatorname{card}K\leq \kappa \}$, for every cardinal number $\kappa$.

*$\{ A\setminus K \mid K\in\mathscr{P}A, \operatorname{card}K< \kappa \}$, for every cardinal number $\kappa$. (added)
Are there any other examples? By the way, are my two classes of examples equal to each other?
 A: Your second type of example is all that there is.

Call the type of a set $X\subseteq A$ the pair $tp(X)=(\kappa, \lambda)$ where $\kappa=\vert X\vert$ and $\lambda=\vert A\setminus X\vert$. Let $F$ be a filter on $A$ which is invariant under self-bijections of $A$.
Claim 1: $F$ is type-invariant: if $X\in F$ and $tp(X)=tp(Y)$, then $Y\in F$. 
Proof: If $tp(X)=tp(Y)$ then there is an automorphism $\alpha$ of $A$ sending $X$ to $Y$; since $F$ is invariant under $\alpha$, this means $Y\in F$. $\Box$
Let the complement spectrum of $F$ be $$CS(F)=\{\lambda: \exists \kappa, X\in F(tp(X)=(\kappa, \lambda))\}.$$ Clearly $CS(F)$ is closed downwards, since $F$ is a filter.
Claim 2: If $F$ contains an element of type $(\kappa_0,\lambda)$, and $\kappa_1+\lambda=\vert A\vert$, then $F$ contains an element of type $(\kappa_1, \lambda)$.
Proof: If $\kappa_1>\kappa_0$, this is easy: pick $X\in F$ of type $(\kappa_0,\lambda)$ and let $Y$ be some superset of $X$ with type $(\kappa_1,\lambda)$ (exercise: such a $Y$ exists).
If $\kappa_1<\kappa_0$, we argue as follows. Let $X\in F$ have type $(\kappa_0, \lambda)$. Let $Y\subseteq A$ have type $(\kappa_0,\lambda)$ with $\vert X\cap Y\vert=\kappa_1$. Since $tp(X)=tp(Y)$, we have $Y\in F$; so $X\cap Y\in F$ since $F$ is a filter. But then we have $tp(X\cap Y)=(\kappa_1,\lambda)$. $\Box$

So let $\theta$ be the smallest cardinal not in $CS(F)$. Then I claim that $$F=\{X\subseteq A: \vert A\setminus X\vert<\theta\}.$$ Why?


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*Suppose $Z\subseteq A$ has $\vert A\setminus Z\vert<\theta$. Then $\vert A\setminus Z\vert\in CS(F)$. But then by Claim 2 above, $F$ contains an element of type $(\vert Z\vert, \vert A\setminus Z\vert)$. But then by Claim 1, since this is the type of $Z$ we have $Z\in F$.

*Suppose $Z\subseteq A$ has $\vert A\setminus Z\vert\ge\theta$. Then $\vert A\setminus Z\vert\not\in CS(F)$. But that means $Z$ can't be in $F$, since otherwise $F$ would have an element of type $(\vert Z\vert, \vert A\setminus Z\vert)$, and hence we'd have $\vert A\setminus Z\vert\in CS(F)$.
