Show that the only endomorphism $\phi$ of $\mathbb{Z}_7 \times \mathbb{Z}_7$ satisfying $\phi^5 = \text{id}$ is the identity. 
Let $\phi: \mathbb{Z}_7 \times \mathbb{Z}_7 \to \mathbb{Z}_7 \times \mathbb{Z}_7$ be a homomorphism such that $\phi^5 = \text{id}$. Show that $\phi$ is the identity.

My attempt: Since $\ker(\phi) \subset \ker(\phi^2)\subset \cdots \subset \ker(\phi^5) = 0$, it follows that $\phi$ is injective. Since $|\mathbb{Z}_7\times \mathbb{Z}_7|$ is finite, $\phi$ is automatically an isomorphism. 
Then $\phi$ should map a subgroup to a subgroup. Consider $\phi(\mathbb{Z}_7 \times 0)$, then it should be mapped to $\mathbb{Z}_7 \times 0$ or $0 \times \mathbb{Z}_7$. Without loss of generality, we consider the former case. In particular, $\phi(1,0)$ should be a generator for $\mathbb{Z}_7 \times 0$. Suppose $\phi(1,0) = (a,0)$. Then $(1,0) = \phi^5(1,0) = \phi^4(a,0) = \phi^3(a^2,0) = \cdots = (a^5,0)$. Thus, $a^5 = 7k+1$ for some $k \in \mathbb{Z}$. By testing $a \in \mathbb{Z}_7$, we see that the only possibility is that $a = 1$. The discussion for $\phi(0,1)$ is similar.
Could anyone help me take a look at my attempt? Is my approach reasonable, and is there any better method to do this problem?
 A: The start is reasonable. However,  there are various other cyclic subgroups, thus I do not see how one could make the rest of the argument work easily.
Instead, I recommend you recall results from Linear Algebra. 
Namely what you know is that the polynomial $X^5 -1$ annihilates $\phi$. 
Thus, the minimal polynomial of $\phi$, which is of degree at most $2$ essentially by Cayley-Hamilton,  divides $X^5 - 1$. 
Now, it remains to check that $X^5 - 1$ only has one factor of degree at most $2$, namely $X-1$, which gives the result, as it means that $X-1$ annihilates $\phi$ meaning precisely that it is the identity.
Note that for the polynomials you have to work over the field with $7$ elements. See Irreducible factors of $X^p-1$ in $(\mathbb{Z}/q \mathbb{Z})[X]$ in case you have difficulty to show the result I mention above. 

Added: Seeing a comment,and thinking about it, it is likely easier to argue in terms of group theory. You have already shown that that $\phi$ is in  $GL_2(F_7)$. That is $\phi$ is an element of that group whose order divides $5$. It thus has order $5$ or $1$. You want to show the latter. To this end show that $GL_2(F_7)$ cannot contain an element of order $5$. This amounts to showing that the order of this group is not divisible by $5$. Now the order of that group is $(7^2 -1)(7^2 -7)$, and the claim follows. 
A: First show that $\phi$ is bijective, as you already did.
Suppose to the contrary that $\phi \neq id$. Consider the number of fixed points of $\phi$. Since they form a subgroup, it should either be $1$ or $7$, so the number of non-fixed points are $48$ or $42$.
As $\phi^5 = id$, you can pack the non-fixed points five in a group, namely, $\{a, \phi(a), \phi^2(a), \phi^3(a), \phi^4(a)\}$. But neither $42$ nor $48$ is a multiple of $5$, which is absurd.
