$G$ is a finite group. $a,b\in G$ and that $\forall n\in \Bbb N$ : $(a^{-1}ba)^{n}=a^{-1}b^{n}a$ can I presume that $(aba^{-1})^{n}=ab^{n}a^{-1}$?

Question

I don't know how to check it.
I can't assume that G is abelian so I can't use the axiom of commutativity.
$(aba^{-1})^{n}=ab^{n}a^{-1}$ is different.

Answer 1

Both equalities hold in any group, by associativity.

$$(a^{-1} b a)^n = (a^{-1} b a)(a^{-1} b a) \cdots (a^{-1} b a) = a^{-1} b^n a$$

$$(aba^{-1})^n = (aba^{-1})(aba^{-1})\cdots(aba^{-1}) = ab^n a^{-1}$$