It is a well known fact that if $a=b \pmod n$ then $a^k = b^k \pmod n$ where $ a, b, n, k $ are integers( $k\neq0.$)
But now I was thinking that is there any special case for which $a^k=b^k\pmod n $ $\implies$ $a=b \pmod n.$
My approach: break $a^k -b^k$ in factors. But I can't proceed further.
One pair of sufficient conditions would be:
$$\gcd(a,n)=\gcd(b,n)=1$$ $$\gcd(k, \phi(n))=1$$
where $\phi$ is Euler's totient function.
This is because the second condition implies that there are integers $u,v$ such that $uk+v\phi(n)=1$ (Bézout's identity), i.e. $uk=1+(-v)\phi(n)$. Then:
$$\begin{array}{rcl}a^k\equiv b^k\pmod n&\implies&a^{uk}\equiv b^{uk}\pmod n\\ &\implies& a\cdot (a^{\phi(n)})^{(-v)}\equiv b\cdot(b^{\phi(n)})^{(-v)}\\&\implies&a\equiv b\pmod n\end{array}$$
because $a^{\phi(n)}\equiv b^{\phi(n)}\equiv 1\pmod n$ (Euler's theorem, applicable because of the first condition).
Exercise: Show that we can also drop one of the conditions $\gcd(a,n)=1$ or $\gcd(b,n)=1$: any one of them is already sufficient.
The question is asking when is the $k$th power map on $\mathbb{Z}_n$ injective. Since $\mathbb{Z}_n$ is finite, this is equivalent to asking when is the map bijective.
If so, then the map must have finite order, so there must be some positive integer $m$ for which $\forall a \in \mathbb{Z} \, a^{k^m} \equiv a \pmod{n}$.
It can be shown that there is an exponent $j \ge 2$ for which $\forall a \in \mathbb{Z}\, a^j \equiv a \pmod{n}$ if and only if $n$ is square-free. Indeed, if $n$ is square-free, then by the Chinese Remainder Theorem, it suffices to check the congruence modulo each prime factor of $n$. Fermat's Little Theorem implies that $a^j \equiv a \pmod{p}$ whenever $p$ is prime and $j \equiv 1 \pmod{p-1}$. From this, it easily follows that the exponent $j$ could be taken to be $\varphi(n)+1$ (or in fact, $\lambda(n)+1$ where $\lambda$ is the Carmichael function). If, on the other hand, $n$ is not square-free (say its prime factorization is $p_1^{e_1}p_2^{e_2}...p_{\omega(n)}^{e_{\omega(n)}}$), then let $a=p_1^{\lceil{e_1/2}\rceil}p_2^{\lceil{e_2/2}\rceil}...p_{\omega(n)}^{\lceil{e_{\omega(n)}/2}\rceil}$. Then, for any $j \ge 2$, $a^j \equiv 0 \not \equiv a \pmod{n}$.
So, if $n$ is not square-free, then the statement holds only for $k=1$. If $n$ is square-free, then the values of $k$ for which the statement holds are exactly those that have a power congruent to $1$ modulo $\lambda(n)$, which are exactly those that are coprime to $\lambda(n)$ (or equivalently, to $\varphi(n)$ since the LCM and product of $p-1$ over the prime factors $p$ of $n$ (which are equal to $\lambda(n)$ and $\varphi(n)$ respectively) have the same distinct prime factors).
