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Let $A,B:C\to C$ be two endofunctors. I wonder if it follows from a natural isomorphism $A\circ A\xrightarrow{\cong}B\circ B$, that $A$ and $B$ are naturally isomorphic.

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No. Consider a ring $R$ and two elements $x,y \in R$ with square zero. Consider the endofunctor $A(M) = xM$ and $B(M) = yM$. Then $A \circ A, B \circ B $ are identically the zero functor, but $A$ and $B$ are generally not isomorphic.

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I assume $A, B$ are acting on $R\text{-Mod}$? –  Qiaochu Yuan Oct 23 '11 at 23:13
    
One can simplify this to: let $C$ be a category with one object $\bullet$, so that it is a complicated way of talking about the monoid $M=\hom_C(\bullet,\bullet)$. Then endofunctors are endomorphisms of $M$, and two endofunctors are naturally isomorphic if they differ by an inner automorphism of $M$. Now, it is easy to show examples of monoids with two endofunctors with the same squares but not differing by an inner automorphism. –  Mariano Suárez-Alvarez Oct 23 '11 at 23:17
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