The role of the rigid relation principle ($RR$) in the Kunen inconsistency Consider the rigid relation ($RR$) principle, i.e.
"every set admits a rigid binary relation", that is,"that for every set $A$ there is a binary relation $R$ on $A$ such that the structure $(A,R)$ is rigid, meaning that it has no nontrivial automorphisms" ("that is, no bijection function $\pi$: $A$$\rightarrow$$A$ such that $aRb$ iff $\pi$$(a)$$R$$\pi$$(b)$, other than the identity function").  (Hamkins and Palumbo, arXiv: 1106.4635v1 [mathLO]).
In that paper, Hamkins and Palumbo prove that $AC$$\Rightarrow$$RR$, and that $ZF+{\lnot}AC+RR$ and $ZF+{\lnot}AC+{\lnot}RR$ are relatively consistent with $ZF$.
Question 1:  What role (if any) does $RR$ play in the proof of the Kunen inconsistency?
Question 2:  Since $ZF+{\lnot}RR$ is relatively consistent with $ZF$, is $NGB+{\lnot}AC+{\lnot}RR$ ($RR$ appropriately defined for $NGB$--call it $RR_{NGB}$) relatively consistent with $NGB$?
Question 3:  Can the Kunen inconsistency be proved in $NBG+{\lnot}AC+{\lnot}RR_{NGB}$?  
 A: Claim: NGB + $\exists j\colon V\to V$ is equiconsistent with NGB + $\exists j\colon V\to V + \neg RR$. 
In order to prove the left to right direction, we will start with a model of NGB + $\exists j\colon V\to V$ and take the symmetric extension that as described in Hamkins and Palumbo's paper. By the results of this paper, in the symmetric extension there is a set that witnesses $\neg RR$. Note that the poset, as well as the groups and the filter, that are used in the symmetric extension are all very low in the cumulative hierarchy (say in $V_{\omega + \omega}$). Therefore, they are fixed by the embedding $j$. 
Let $W$ be the symmetric extension and let's extend $j$ to $\tilde{j}\colon W\to W$. 
We construct $\tilde{j}$ as in the proof of Levy-Solovay's theorem. For a hereditarily symmetric name $\dot{x}$, let's define $\tilde{j}(\dot{x}^G) := j(\dot{x})^G$. 
Since $j$ does not move the forcing notion, $\mathbb{P}$, the groups of automorphisms of the forcing, $\mathcal{G}$, and the filter, $\mathcal{F}$, $j(\dot{x})$ is a hereditarily symmetric name (being hereditarily symmetric is a first order property with parameters $\mathbb{P}, \mathcal{G}, \mathcal{F}$), and therefore $j(\dot{x})\in W$. $\tilde{j}$ is elementary, by exactly the same argument of Levi-Solovay: using the forcing theorem (every statement in the generic symmetric extension is forces by some condition in the generic filter), and the fact that $j$ doesn't move the forcing or the conditions, $p\Vdash \varphi(\dot{x})$ iff $j(p) = p\Vdash \varphi(j(\dot{x}))$, by the elementarity of $j$ in $V$.    
Moreover, the class $\dot{j} = \{(\dot{x}, j(\dot{x})) \mid \dot{x} \text{ is hereditarily symmetric name}\}$ (when writing the pairs as canonical names of pairs) is a hereditarily symmetrical name for a class (it is fixed by itself by any automorphism in the group, and every member of it is a pair of hereditarily symmetric names). So $\tilde{j}$ is a class of $W$. 
A: Question 2 is an immediate consequence of the theorems in my paper with Justin. 


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*J. D. Hamkins and J. Palumbo, The rigid relation principle, a new weak choice principle, Mathematical Logic Quarterly, vol. 58, iss. 6, pp. 394-398, 2012.


Specifically, in that paper, we prove in theorem 5 that if there is a model of $\text{ZF}$, then there is a model of $\text{ZF}+\neg\text{AC}+\neg\text{RR}$. If one augments this model with the collection of definable classes (definable from parameters), then one arrives at a model of $\text{NGB}$, which has exactly the same sets as the original model, and therefore also $\neg\text{AC}+\neg\text{RR}$. So $\text{ZF}$ is equiconsistent with $\text{NGB}+\neg\text{AC}+\neg\text{RR}$. Doing the same thing with the model of theorem 6 shows that it is also equiconsistent with $\text{NGB}+\neg\text{AC}+\text{RR}$, as well as with $\text{NGB}+\text{global choice}$. 
Meanwhile, I don't know whether one can prove the Kunen inconsistency using merely $\text{RR}$ in place of $\text{AC}$, but I find this to be a very interesting idea. (Note, however, that this is not actually what you asked, although I find it to be in the same spirit as what you asked.) 
