= Solution
A <measurable cardinal> is an uncountable cardinal $\kappa$ carrying a <nonprincipal ultrafilter> $U$ on $\kappa$ that is $\kappa$-complete: intersections of fewer than $\kappa$ members of $U$ remain in $U$. An <elementary embedding> $j:M\to N$ preserves truth of every <first-order formula>, with its parameters carried along by $j$.
The identity is always an elementary embedding of the universe into itself. The substantive question concerns a nonidentity embedding. In <ZFC>, the <Kunen inconsistency theorem> gives
$$
\boxed{\text{There is no nontrivial elementary embedding }j:V\to V.}
$$
Here, as usual for class embeddings, restrictions to sets and the recursively defined critical sequence are available. We give the stationary-set proof, including its partition ingredient.
Suppose such a $j$ existed. It has a least moved <ordinal> $\kappa$, its <critical point of an elementary embedding>. To see why some <ordinal> must move, if all <ordinals> were fixed, induction on rank would fix every set: membership of $z$ in $j(x)$ is equivalent to membership of $z$ in $x$ once every lower-rank $z$ is fixed. A rank segment containing a moved set would then be fixed pointwise, a contradiction. The critical point is an uncountable cardinal. All finite <ordinals> and $\omega$ are definable and fixed. If $\kappa$ had <cardinality> $\mu<\kappa$, a bijection $f:\mu\to\kappa$ would be carried to a surjection $j(f):\mu\to j(\kappa)$ whose range still consists exactly of the fixed <ordinals> below $\kappa$, impossible since $j(\kappa)>\kappa$.
Put $\kappa_0=\kappa$, $\kappa_{n+1}=j(\kappa_n)$ and $\lambda=\sup_{n<\omega}\kappa_n$. Applying $j$ to this <Kunen critical sequence> shifts it by one, so $j(\lambda)=\lambda$. Elementarity also gives $j(\lambda^+)=\lambda^+$; write $\theta=\lambda^+$.
We need a <stationary partition at a successor cardinal>. If $S\subseteq\lambda^+$ is stationary, choose bijections $e_\beta:\lambda\to\beta$ for $\lambda\leq\beta<\lambda^+$. For each $\xi<\lambda^+$, on the stationary tail $\{\beta\in S:\beta>\max(\lambda,\xi)\}$ the function $\beta\mapsto e_\beta^{-1}(\xi)$ is regressive. The <Fodor lemma> supplies a stationary fiber $E_\xi$ with constant value $i_\xi<\lambda$. Some value $i$ occurs for $\lambda^+$ many $\xi$, since a union of $\lambda$ sets each of size at most $\lambda$ cannot have size $\lambda^+$. For distinct such $\xi$, their fibers are disjoint: $e_\beta(i)$ cannot equal two different <ordinals>. These fibers yield $\lambda^+$ disjoint stationary subsets; include any leftover points in one cell. Grouping cells also gives a partition into any nonzero number at most $\lambda^+$.
For completeness, the <Fodor lemma> used here follows from diagonal intersections of clubs. If a regressive function $f:S\to\theta$ had no stationary fiber, choose a <club set> $C_\xi$ disjoint from each fiber. The diagonal intersection $\{\beta<\theta:\forall\xi<\beta\ (\beta\in C_\xi)\}$ is club by regularity and closure. A point $\beta$ in its intersection with $S$ would belong to $C_{f(\beta)}$, contradicting the choice of that club.
The set $S=\{\beta<\theta:\operatorname{cf}(\beta)=\omega\}$ is stationary: inside any club choose a strictly increasing countable sequence, whose supremum still lies below the regular uncountable $\theta$ and belongs to the club. Partition it as $\langle S_\alpha:\alpha<\kappa\rangle$. Its image $\langle T_\alpha:\alpha<j(\kappa)\rangle$ is another stationary partition of the same $S$.
The set $C=\{\beta<\theta:j\mathbin{``}\beta\subseteq\beta\}$ is club. Closure is immediate. To get a member above any given <ordinal>, repeatedly bound the images of all earlier <ordinals> and take a countable supremum; regularity of $\theta$ keeps each bound below $\theta$. Choose $\beta\in C\cap T_\kappa$. Since $\operatorname{cf}(\beta)=\omega$, a countable cofinal sequence in $\beta$ and $j(\omega)=\omega$ give $j(\beta)=\beta$: the image sequence remains below $\beta$ by closure under $j$, and $j(\beta)\geq\beta$.
But $\beta\in S_\alpha$ for some $\alpha<\kappa$. Applying $j$ and using $j(\alpha)=\alpha$ gives $\beta=j(\beta)\in T_\alpha$, contradicting its membership in the disjoint cell $T_\kappa$. This proves the assertion. The use of <axiom of choice> in the cardinal and stationary-partition arguments matters; the conclusion here is explicitly in ZFC.
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