Show that the only endomorphism $\phi$ of $\mathbb{Z}_7 \times \mathbb{Z}_7$ satisfying $\phi^5 = \text{id}$ is the identity.

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.

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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.

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.