Fix $k\in \mathbb{Z}$ and $g\in G$, a group. The equation $x^k=g$ has the same number of solutions in $G$ as does $y^k=\phi(g)$ in $G^\prime$ where $\phi:G\to G^\prime$ is an isomorphism.
My attempt.
Let $x\in G^\prime$ such that $x^k=\phi(g)$, since isormophisms are surjective we know there exists an element $h\in G$ such that $\phi(h)=x$, thus $\phi(h)^k=\phi(g)$. And since isormorphisms are injective, then $\phi(h)^k=\phi(h^k)=\phi(g)$ implies $h^k=g$. Thus every solution to $x^k=\phi(g)$ in $G^\prime$ gives exactly 1 solution to $x^k=g$ in $G$.
Suppose $h\in G$ is a solution so that $h^k=g$ then $\phi(h^k)=\phi(h)^k=\phi(g)$, so $\phi(h)$ is a solution to $x^k=\phi(g)$ in $G^\prime$. Since isomorphisms are injective there is only 1 element $y\in G$ such that $\phi(y)=\phi(h)$ thus every solution to $x^k=g$ gives exactly 1 solution to $x^k=\phi(g)$ in $G^\prime$.
Since each solution in either group gives exactly 1 solution in the other, the number of solutions is preserved and thus they have the same number of solutions.
I think I have the idea but I'm not sure my proof really explains the reasoning too well. Also is isomorphism necessary? Or is injective homomorphim sufficent?
Answer was provided in comments but I had the thought that simply by injectivity $\phi$ will map solutions in $G$ to solutions in $G^\prime$ as $x^k=g \iff \phi(x)^k=\phi(g)$. Similarly it will also map elements which are not solutions in $G$ to elements which are not solutions in $G^\prime$ as $x^k\not= g \iff \phi(x)^k\not= \phi(g)$.
This however isn't sufficient as it does not account for solutions which are present in the codomain but are not in the image of $\phi$. Thus surjectivity is necessary.