= Solution
If $y=gxg^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=gx^kg^{-1}$ for every positive integer $k$. Hence $y^k$ is the <identity element> exactly when $x^k$ is, proving that <conjugate group elements> have the same <order of a group element>. For <permutations>, the sign <group homomorphism> similarly gives
$$
\operatorname{sgn}(gxg^{-1})=\operatorname{sgn}(g)\operatorname{sgn}(x)\operatorname{sgn}(g)^{-1}=\operatorname{sgn}(x).
$$
Thus <conjugate permutations> have the same <sign of a permutation>.
The converse is false: <permutation order and sign do not determine conjugacy>. In the <symmetric group> $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gxg^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their <cycle types> $3\,1^3$ and $3^2$ differ. \b[Equal order and sign are necessary, but not sufficient, for conjugacy.]
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