Yes, assuming $X$ is reduced (so that its absolute Frobenius endomorphism is schematically dominant); this really is a formal consequence of the usual descent formalism (despite whatever red herring appears with Frobenius maps not being isomorphisms).

As a warm-up, consider the case $n=1$ (so $X^{(q)} = X$ tautologically and ${\rm{Fr}}$ is thereby identified with the absolute $q$-Frobenius): for this you're asking that $i \circ {\rm{Fr}} = {\rm{Fr}}$, which is to say that $i = {\rm{id}}$ since ${\rm{Fr}}$ is schematically dominant, which is indeed the only possible Galois descent datum when $n=1$. So below we will assume $n>1$ to make things interesting.

By the universal functoriality of the relative $q$-Frobenius, the $n$th power of that composition is the same as $X \rightarrow X^{(q^n)} \simeq X$ where the first step is the relative $q^n$-Frobenius and the second step is the composition $i \circ i^{(q)} \circ \dots i^{(q^{n-1})}$ (notation meaningful as written since $n>1$). But the relative $q^n$-Frobenius of $X$ is identified with its absolute Frobenius via the canonical isomorphism $X \simeq X^{(q^n)}$ defined via how $X$ is equipped with a structure of $\mathbf{F}_{q^n}$-scheme. Hence, again by the dominance condition (using reducedness of $X$) we can "cancel" as when $n=1$ to see that the criterion in your question is exactly the condition that this latter composite isomorphism coincides with the *canonical* isomorphism $X^{(q^n)} \simeq X$ already mentioned.

Now all of the usual confusion that permeates every discussion of absolute and relative Frobenius maps has been eliminated. And to clarify matters let's get rid of all mention of finite fields by instead considering a cyclic extension of fields $K/k$ of degree $n>1$ with a chosen generator $\gamma$ of ${\rm{Gal}}(K/k)$ and an arbitrary $K$-scheme $X$ (no need for reducedness anymore) equipped with a $K$-isomorphism $i:\gamma^{\ast}(X) \simeq X$. Any $K/k$-descent datum on $X$ gives rise to such an $i$ which moreover uniquely determines it (by the cocycle condition), and we can ask to characterize those $i$ which arise in that way. Namely, a necessary condition is that the composite isomorphism
$$i \circ \gamma^{\ast}(i) \circ \dots \circ (\gamma^{n-1})^{\ast}(i): (\gamma^n)^{\ast}(X) \simeq X$$
coincides with the canonical such isomorphism (using that $\gamma^n = {\rm{id}}_K$).

Let's make one final maneuver, to turn this into entirely a question about actual automorphisms of $X$ (though only over $k$, not $K$), exactly in the spirit of the elegant discussion of Galois descent as a special case of Grothendieck's descent formalism in Example B in 6.2 of the book "Neron Models". Namely, by the universal property of fiber products a $K$-isomorphism $i: \gamma^{\ast}(X) \simeq X$ is the "same" as a scheme isomorphism $i':X \simeq X$ over the *inverse* of the automorphism ${\rm{Spec}}(\gamma)$ of ${\rm{Spec}}(K)$ (so $i'$ is over $k$), or in other words $i^{-1}$ corresponds to an automorphism of the scheme $X$ over ${\rm{Spec}}(\gamma)$. In this way we see that the condition we are imposing on that $n$-fold composition is precisely the statement that $i'$ has $n$-fold composition equal to the identity (or same for ${i'}^{-1}$ if you prefer). That in turn is precisely saying that we are giving a $k$-action on $X$ by a cyclic group of order $n$, or more specifically an action by ${\rm{Gal}}(K/k)$ covering its action on ${\rm{Spec}}(K)$ over ${\rm{Spec}}(k)$ (I leave to you the pleasure of getting left-actions and right-actions straightened out by thinking about the analogue of these matters with a possibly non-abelian Galois group). Referring again to the exposition given in "Neron Models" as cited above, we are done.

QED