= Solution
A <measurable cardinal> is an uncountable <cardinal> $\kappa$ carrying a <kappa-complete> <nonprincipal ultrafilter> $U$ on $\kappa$. An <elementary embedding> $j:M\to N$ preserves all <first-order formulas> with parameters:
$$
M\models\varphi(a_1,\ldots,a_r)\quad\Longleftrightarrow\quad
N\models\varphi(j(a_1),\ldots,j(a_r)).
$$
For a measurable $\kappa$, the <Łoś theorem> gives an elementary map from the universe into its <ultrapower>. Countable completeness makes that ultrapower well founded, so the <Mostowski collapse> gives a transitive target $M$ and an <ultrapower embedding> $j:V\to M$. The least moved ordinal, its <critical point of an elementary embedding>, is $\kappa$: every function into an ordinal below $\kappa$ is constant on a $U$-large set by <kappa-completeness>, whereas the identity function on $\kappa$ represents an ordinal above every constant ordinal below $\kappa$ and below the image of $\kappa$. Conversely a suitably available embedding with critical point $\kappa$ defines the measure $U=\{X\subseteq\kappa:\kappa\in j(X)\}$.
The identity $V\to V$ is elementary. \b[In ZFC there is no nonidentity elementary class embedding $j:V\to V$], by the <Kunen inconsistency theorem>. Here the embedding is a class for which restrictions to sets and the critical sequence are available, as in the usual class-theoretic formulation. The <axiom of choice> is part of this conclusion; it is not a claimed contradiction for the choice-free Reinhardt-cardinal hypothesis.
Here is the stationary-set proof. A nonidentity $j$ has a least moved ordinal $\kappa$; if all ordinals were fixed, induction on rank would fix every set. Put $\kappa_0=\kappa$, $\kappa_{n+1}=j(\kappa_n)$ and $\lambda=\sup_{n<\omega}\kappa_n$. Since $j$ fixes $\omega$ and shifts this sequence, $j(\lambda)=\lambda$. It therefore fixes the regular <successor cardinal> $\delta=\lambda^+$. The <stationarity of ordinals of prescribed cofinality> gives the stationary set
$$
S=\{\beta<\delta:\operatorname{cf}(\beta)=\omega\}.
$$
Use the <stationary partition at a successor cardinal> to split $S$ into $\kappa$ disjoint stationary pieces $\langle S_\alpha:\alpha<\kappa\rangle$, merging pieces if necessary. Applying $j$ gives a partition $\langle T_\alpha:\alpha<j(\kappa)\rangle$ of the same $S$ into stationary pieces. The additional piece $T_\kappa$ is stationary.
The closure points
$$
C=\{\beta<\delta:(\forall\gamma<\beta)\ j(\gamma)<\beta\}
$$
form a <club set>. For unboundedness, repeatedly close a bound under the values of $j$ below it and take the supremum of the resulting countable sequence; regularity of $\delta$ keeps it below $\delta$. Closure under limits is immediate. Choose $\beta\in C\cap T_\kappa$ and an increasing sequence $\langle\beta_n:n<\omega\rangle$ cofinal in $\beta$. Then
$$
j(\beta)=\sup_{n<\omega}j(\beta_n)\le\beta,
$$
while every ordinal satisfies $j(\beta)\ge\beta$, so $j(\beta)=\beta$. There is an $\alpha<\kappa$ with $\beta\in S_\alpha$. Since $j(\alpha)=\alpha$, elementarity gives $\beta=j(\beta)\in j(S_\alpha)=T_\alpha$, contrary to disjointness from $T_\kappa$.
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