Which filters on a set $A$ are invariant under every bijection $A\leftrightarrow A$?
I found the following examples of invariant filters:
$\{ A\setminus K \mid K\in\mathscr{P}A, \operatorname{card}K\leq \kappa \}$, 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?
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?
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 $\vert A\setminus Z\vert\in CS(F)$.