Let $S_3$ be a set of all permutations of elements in $\{1,2,3\}$. Prove that there doesn't exist f $\in S_3$ where $\{f,f^2,f^3,f^4,f^5,f^6\} = S_3$.
Where $f^n = f \circ f \circ \:... \circ \:f$ $n$ times.
The proof in the answer sheet goes about proving this by showing that $f \circ g \neq g \circ f$ for any $g \in S_3$. However my reasoning was like so:
All permutations of $\{1,2,3\}$ are of order $2$ or $3$, and therefore there are at most $3$ unique permutations for any $f^n$ where $f \in S_3$. However the number of unique permutations of $\{1,2,3\}$ is $3! = 6$.
Therefore $\{f,f^2,...,f^n\}$ does not contain all permutations of $\{1,2,3\}$, and so $\{f,f^2,...,f^n\} \neq S_3$ for any $f$ and any $n$.
Is my reasoning correct?
That looks reasonable to me; such an $f$ would need to have order 6, and you can show there is no such $f$.
There is a very minor flaw in your argument: it is not true that all permutations of $\{1,2,3\}$ have order 2 or 3. There is one exception...