The automorphism group of a cyclic group $\mathbb{Z}_n$ has an order given by Euler's totient function $\phi(n)$, where $\phi(n)$ is the number of generators of $\mathbb{Z}_n$. The answer in this question says that this is because generators must map to generators, presumably as an isomorphism (and hence an automorphism) preserves element order.
Consider some groups such all non-identity elements have equal order:
The Klein-4 group $\mathbb{Z}_2 \times \mathbb{Z}_2$ has the identity element and 3 elements of order 2. Taking permutations of these 3 elements gives an automorphism group of order $3! = 6$, which is the correct order.
The Elementary Abelian group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ has the identity element and 7 elements of order 2. Taking permutations of these 7 elements gives an automorphism group of order $7! = 5040$, but the correct order is $168$.
The group $\mathbb{Z}_3 \times \mathbb{Z}_3$ has the identity element and 8 elements of order 3. Once again taking permutations give the automorphism group an order of $8! = 40320$, but the correct order is $48$.
Is there a way to compute the order of the automorphism groups from element order alone?
Some examples use the result ${\rm Aut}(C_p\times C_p)\simeq{\rm GL}_2(\mathbb Z_p)$, but interested to see if it can be computed using the element order alone.