= Solution
Each <endomorphism> has three distinct <eigenvalues> in a three-dimensional <complex vector space>. Choose a nonzero <eigenvector> for each <eigenvalue>; these vectors are <linearly independent> and hence form an <eigenbasis>. Both operators are therefore <diagonalizable> with diagonal matrix $D=\operatorname{diag}(1,2,3)$.
Their <characteristic polynomials> are
$$
\boxed{\chi_S(t)=\chi_T(t)=(t-1)(t-2)(t-3).}
$$
If a <polynomial> $p$ annihilates either operator, applying it to each <eigenvector> gives $p(1)=p(2)=p(3)=0$. Thus its <minimal polynomial> must contain all three factors. Conversely, their product annihilates the diagonal matrix, so
$$
\boxed{m_S(t)=m_T(t)=(t-1)(t-2)(t-3).}
$$
Finally, send an ordered <eigenbasis> of $S$ to the correspondingly ordered <eigenbasis> of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence \b[$T=CSC^{-1}$: the two endomorphisms are conjugate].
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