This is a supplementing answer to aws' answer. In this answer I sketch the a class-symmetric extension in which there is a proper class of Russell sets, and there is no set of Russell cardinals.

(Recall that a Russell set is a set which can be partitioned into countably many pairs, such that no infinite family of those pairs admits a choice function.)

We work in $\sf ZFC+GCH$, where $\sf GCH$ is assumed for simplicity. $\DeclareMathOperator{\dom}{dom} \DeclareMathOperator{\sym}{sym} \DeclareMathOperator{\fix}{fix} \newcommand{\PP}{\Bbb P} \newcommand{\cG}{\mathcal G} \newcommand{\HS}{\mathsf{HS}} \newcommand{\ZF}{\mathsf{ZF}} \newcommand{\cF}{\mathcal F} \newcommand{\id}{\operatorname{id}} \newcommand{\tup}[1]{\langle #1\rangle}$

If $\PP$ is a forcing, and $\{\dot x_i\mid i\in I\}$ is a set (or class) or $\PP$-names, we define $\{\dot x_i\mid i\in I\}^\bullet$ to be the $\PP$-name $\{\tup{1_\PP,\dot x_i}\mid i\in I\}$. This extends to ordered pairs, and naturally to sequences.

##1 Localized failure

Fix a regular cardinal $\kappa$. Let $\PP_\kappa$ be the following presentation of adding a Cohen subset to $\kappa$. We mimic the construction of Cohen's second model.

A condition in $\PP_\kappa$ is a partial function $p\colon\omega\times 2\times\kappa\to2$, such that $|\dom p|<\kappa$.

We define the following names, for $(n,i,m)\in\omega\times2\times\omega$:

• $\dot x_{n,i,m}=\{\tup{p,\check\alpha}\mid p(n,i,m,\alpha)=1\}$,
• $\dot X_{n,i}=\{\dot x_{n,i,m}\mid m\in\omega\}^\bullet$, and
• $\dot S_n=\{\dot X_{n,i}\mid i<2\}^\bullet$.

Next we define the automorphism group $\cG_\kappa$, to be permutations $\pi$ of $\omega\times2\times\omega$ moving only finitely many points and satisfying that if $\pi(n,i,m)=(n',i',m')$, then $n=n'$; and for all $n$, either $i=i'$ or $i'=1-i$. It will be clearer to understand, once we see how these act on the names defined above:

• If $p\in\PP_\kappa$, then $\pi p(\pi(n,i,m),\alpha)=p(n,i,m,\alpha)$. This is the standard way this action is defined. It extends to $\PP_\kappa$-names recursively: $\pi\dot x=\{\tup{\pi p,\pi\dot y}\mid\tup{p,\dot y}\in\dot x\}$.

• $\pi\dot x_{n,i,m}=\dot x_{\pi(n,i,m)}$,

• $\pi\dot X_{n,i}=\dot X_{n,i'}$ if there are some $m$ and $m'$ such that $\pi(n,i,m)=\pi(n,i',m')$, and

• $\pi\dot S_n=\dot S_n$.

In other words, for every pair $A_n$, we decide whether or not we switch the elements in that pair, and separately, we may permute the elements of each set in the pair.

Finally, we take $\cF_\kappa$ to be the normal filter of subgroups generated by fixing finitely many points. Namely, if $E\subseteq\omega\times2\times\omega$ is finite, $\fix(E)$ is the group of all automorphisms which fix all the points in $E$ pointwise; and $\cF_\kappa$ is generated by $\fix(E)$ for $E$ finite.

We say that a name is symmetric if $\{\pi\mid \pi\dot x=\dot x\}\in\cF_\kappa$; and hereditarily symmetric if also every name which appears inside $\dot x$ is hereditarily symmetric.

We denote by $\HS$ the class of hereditarily symmetric names. If $G$ is a generic filter, then $\HS^G=\{\dot x^G\mid\dot x\in\HS\}$ is a model of $\ZF$. Let $M$ denote $\HS^G$, and we omit the dots to indicate the interpretations of the names, e.g. $\dot A_n^G$ is $A_n$ and so on.

Standard arguments show that:

1. Each $\dot x_{n,i,m}$ is symmetric (and thus hereditarily symmetric),
2. Each $\dot X_{n,i}$ is symmetric (and thus ...),
3. The sequence $\tup{\dot A_n\mid n<\omega}^\bullet$ is hereditarily symmetric.
4. The name $\dot A=\{\dot X_{n,i}\mid(n,i)\in\omega\times 2\}^\bullet$ is hereditarily symmetric.

So all these will then be in $M$. Moreover,

5. For all $n$ and $i$, $X_{n,i}$, in $M$ is a Dedekind-finite set.
6. $A$ is a Russell set in $M$. Specifically, $\{A_n\mid n<\omega\}$ is a partition witnessing that.

Global failure

Let $D$ be a class of regular cardinals. For example, all regular cardinals. For every $\kappa$ in $D$, note that $\PP_\kappa$ is $\kappa$-closed. Let $\PP$ be the Easton product of the $\PP_\kappa$'s. We then define an Easton support product of the groups and filters.

The class of hereditarily symmetric names of this system will include all the Russell sets we added by each $\PP_\kappa$. So it remains to show that there is no set of Russell sets which catch all of them up to equi-cardinality.

But then we get this quite easily. Any set of Russell sets will be added by some condition, and then some large enough $\kappa$ will add a new Russell set.

You can find details of a similar construction here:

Karagila, Asaf, Embedding orders into the cardinals with $\mathsf {DC}_{\kappa} $, Fundam. Math. 226, No. 2, 143-156 (2014). ZBL1341.03068.