= Solution
Let $n=\operatorname{ord}(a)$, which is finite. A <group homomorphism> preserves powers and the identity, so
$$
f(a)^n=f(a^n)=f(e)=e'.
$$
If $r=\operatorname{ord}(f(a))$, divide $n$ by $r$: $n=qr+s$ with $0\leq s<r$. Then $e'=f(a)^n=f(a)^s$, and minimality of $r$ forces $s=0$. Thus
$$
\boxed{\operatorname{ord}(f(a))\mid\operatorname{ord}(a),\qquad
\operatorname{ord}(f(a))\leq\operatorname{ord}(a)}.
$$
If $f$ is a <group isomorphism>, apply the same argument to its inverse: $\operatorname{ord}(a)\leq\operatorname{ord}(f(a))$. The two inequalities yield
$$
\boxed{\operatorname{ord}(f(a))=\operatorname{ord}(a)}.
$$
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