For which $n\in \Bbb N$,
$$a\equiv b,a'\equiv b'\quad \text{implies} \quad \gcd(a,a')\equiv \gcd(b,b'), \text{lcm}(a,a')\equiv \text{lcm}(b,b')$$ all mod $n$.
For $n=2$ it is true.
For which $n\in \Bbb N$,
$$a\equiv b,a'\equiv b'\quad \text{implies} \quad \gcd(a,a')\equiv \gcd(b,b'), \text{lcm}(a,a')\equiv \text{lcm}(b,b')$$ all mod $n$.
For $n=2$ it is true.
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This is trivially true when $n=1$, and easily verified to be true when $n=2$. Consider $n \geq 3$. Consider
\begin{equation} 1 \equiv (n-1)^2 \pmod{n}, n-1 \equiv n-1 \pmod{n} \\ \gcd(1, n-1)=1 \not \equiv -1 \equiv \gcd((n-1)^2, n-1) \pmod{n} \end{equation}
Therefore it is only true for $n=1, 2$.