Isomorphism $f$ preserves the order of an element? Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.

Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?

 A: "Isomorphism" means that the elements of $G$ obtain new names taken from the set $G'$, but  everything operationwise stays the same. It is therefore innate in the very essence of "isomorphism" that the claim you are told to prove is true. Any "formal proof" would make the claim less believable.
A: More generally, if $f:G  \rightarrow G'$ is a group homomorphism, then $o(f(x))$ divides $o(x)$ simply because $o(x)=m$ implies $e'=f(e)=f(x^m)=f(x)^m$. In particular, $o(f(x)) \le o(x)$.
When $f$ is an isomorphism with inverse $g$, then we get $o(x) = o(gf(x))\le o(f(x)) \le o(x)$ and so $o(f(x)) = o(x)$.
A: Put $y = f(x)$. If $o(x) = n$, then we have $y^n = f(x)^n = f(x^n) = 1$, so that $o(y)\le n$. If $m < n$ and $1 = y^m = f(x)^m = f(x^m)$, then applying $f^{-1}$ to both sides of this equation and noting that isomorphisms send $1\mapsto 1$, we have $1 = x^m$, absurd.
A: Yes.  Isomorphisms preserve order.  In fact, any homomorphism $\phi$ will take an element $g$ of order $n$ to an element of order dividing $n$, by the homomorphism property.  Now since an isomorphism has an inverse which is also a homomorphism, the claim follows.  For we get that $n|o(\phi(g))$.
A: The claim holds also for infinite $\langle g\rangle$. In fact, by the injectivity, $\langle g\rangle\cong f(\langle g\rangle)$; moreover, $f(\langle g\rangle)\le\langle f(g)\rangle$ generally holds for any homomorphism. If $\langle g\rangle$ is infinite, then $\langle g\rangle\cong f(\langle g\rangle) \cong \Bbb Z$, and hence $\langle f(g)\rangle\cong \Bbb Z$. Therefore, $|\langle g\rangle|=|\langle f(g)\rangle|$.
