I was given this exercise:
Let $f: G \to H$ be an isomorphism. Prove that $|f(x)|=|x|$. Deduce that any two isomorphic groups have same number of elements of order $n$.
In the previous exercise, I had proven the trivial result $f(x^n)=f(x)^n$ and at first glance I thought this was enough to prove that elements of a certain order map to elements of same order, because:
Let $$\begin{align} |g|=n , g\in G &\implies f(g^n) = 1_H \\ &\implies f(g)^n = 1_H. \end{align}$$
The question I have is whether this proof is enough, because I guess I can't guarantee that $n$ is the smallest to which $h^n = 1$.
You have shown that $|f(g)|\bigm ||g|$, but you also know that if $f$ is an isomorphism, then so is $f^{-1}$. You can then repeat the argument with $f^{-1}$ on $f(g)$ to show that $|g|\bigm ||f(g)|$, which gives equality of orders (since if two numbers divide each other they are equal up to a factor of $\pm1$, and both are positive here so this extra factor doesn't matter).
Note: I'm using the fact that if $g^m=1$ for some $m$, then $|g|\bigm |m$. You can see this by using the division algorithm to write $m=p|g|+r$ where $r<|g|$.