The comments are on point here; I'm going to write them up in more detail.
In general, $$(a^{-1}ba)^n = a^{-1}b^na$$ for all $a, b,$ and $n$. To see this, consider $n=2$ as an example:
$$\begin{align}
(a^{-1}ba)^2 & = (a^{-1}ba)(a^{-1}ba) \\
& = (a^{-1}b)(aa^{-1})(ba) \\
& =(a^{-1}b)(ba) \\
& = a^{-1}b^2a
\end{align}
$$
and in general $(a^{-1}ba)^n = (a^{-1}ba)(a^{-1}ba)\dots (a^{-1}ba)$, the $aa^{-1}$ pairs in the middle all cancel out, and the whole thing telescopes down to $a^{-1}b^na$.
We could of course take $c=a^2$ and then because $$(c^{-1}bc)^n = c^{-1}b^nc,$$ therefore $$(a^{-2}ba^2)^n = a^{-2}b^na^2.$$
This operation of combining two elements $a$ and $b$ in this way is extremely important in group theory. It's called conjugation of $b$ by $a$. To start getting an intuition for it, think about a square. It has several symmetries. You can flip it horizontally, or vertically, or diagonally; you can rotate it by various amounts. There's a certain intuitive way in which the horizontal and vertical flips are like each other, and not like the other symmetries. For example, these two flips are the only symmetries of the square that leave two edges fixed in place but move the other two edges. The rotations and the diagonal flips move all four edges. The identity element leaves all four edges in place.
In fact the horizontal and vertical flip “look” the same in the following sense: If you give the square a horizontal flip, while your head is tilted to one side, it looks to you like you're giving it a vertical flip! A horizontal flip is just
- Turn your head sideways
- Vertical flip
- Turn your head back
Or if you prefer, you can turn the square instead of turning your head:
- Rotate the square 90° clockwise
- Vertical flip
- Rotate back
Letting $r$ represent the 90° rotation, and $V$ and $H$ the flips, we have $$H = r^{-1}Vr$$
which is a true equation in the symmetry group of the square, which is called $D_8$. (Or sometimes $D_4$, notation isn't quite standard.)
We say that $H$ and $V$ are conjugate. They are similar in this special way. They are not conjugate to the diagonal flips, which are different, although the diagonal flips are conjugate to each other.
Every group can be partitioned into these families of ‘conjugate’ symmetries. In a square, the horizontal and vertical flips are conjugate, the two diagonal flips are conjugate, the two 90° rotations are conjugate, and the identity and the 180° rotation are not conjugate to anything (except to themselves).
A normal subgroup of a group $G$ is one that contains only complete families from $G$ — if it contains an element $x$ it also contains all the elements of $G$ that are conjugate to $x$. If a normal subgroup of the square contains the horizontal flip, it must also contain the vertical flip. If it contains one of the diagonal flips it must also contain the other. And this is in some sense the reason why normal subgroups are important. If a normal subgroup contains some symmetry, it contains all the symmetries that are “like” that one in this way.
The equation you were asking about, that $(a^{-1}ba)^n = a^{-1}b^na$, should make more intuitive sense now. It says that if you are going to repeat several times the operation of turning your head, then doing $b$, then turning your head back, you can skip some of the head-turning and just turn your head once, do operation $b$ repeatedly, and turn your head back at the end.
For another discussion of this see Why this $\sigma \pi \sigma^{-1}$ keeps appearing in my group theory book? (cycle decomposition). One of the answers there discusses the idea in connection with the symmetries of a sphere.
