The statement is true. The condition that $M'$ is a DFA implies that for every state $r \in Q$ and every symbol $c \in \Sigma$, there is exactly one $q \in Q$ such that $\delta(q, c) = r$. This means that for each $c \in \Sigma$, the function $\delta_c : Q \to Q$ defined by $\delta_c(q) = \delta(q, c)$ is a bijection, i.e., a permutation of $Q$. Consequently, the transformation monoid of $M$ is a group of permutations on $Q$.

Now, consider $L(M)$. If $F = \emptyset$, then $L(M) = \emptyset$. If $F \neq \emptyset$, then if no final state is reachable from $q_0$, $L(M) = \emptyset$. If some final state $f \in F$ is reachable, say $\delta(q_0, w) = f$ for some $w \in \Sigma^*$, then since the transformation monoid is a finite group and $\Sigma^*$ is infinite, there are infinitely many words that map $q_0$ to $f$ under the transformation, so $L(M)$ is infinite. Thus, in all cases, $L(M)$ is either empty or infinite.

The statement is True.

Here is a proof.

Let $M=(Q,\Sigma,\delta,q_0,F)$ be a DFA. The automaton $M'=(Q,\Sigma,\delta',q_0,F)$ is obtained by reversing all transitions, i.e., $\delta'(r,c)=q$ if and only if $\delta(q,c)=r$. We are given that $M'$ is also a DFA. We will show that this condition implies that either $L(M)$ is empty or $L(M)$ is infinite.

1. Analyzing the condition on $M'$

For $M'$ to be a DFA, its transition function $\delta': Q \times \Sigma \to Q$ must be well-defined. This means that for every state $r \in Q$ and every symbol $c \in \Sigma$, there must exist exactly one state $q \in Q$ such that $\delta'(r,c) = q$.

Using the definition of $\delta'$, this condition translates to the properties of the original transition function $\delta$:
For every state $r \in Q$ and symbol $c \in \Sigma$, there is exactly one state $q \in Q$ such that $\delta(q,c) = r$.

This property means that for any fixed symbol $c \in \Sigma$, the function $\delta_c: Q \to Q$ defined by $\delta_c(q) = \delta(q,c)$ is a permutation of the finite set of states $Q$. A function on a finite set is a permutation if and only if it is a bijection (both injective and surjective). The condition directly states that $\delta_c$ is surjective for every $c$ (every state $r$ is an image), and since $Q$ is finite, this also implies it is injective.

2. Consequences for longer strings

Let $w = c_1c_2\dots c_k$ be any non-empty string in $\Sigma^+$. The extended transition function for $w$, denoted $\hat{\delta}_w(q) = \hat{\delta}(q,w)$, is the composition of the functions for each symbol: $\hat{\delta}_w = \delta_{c_k} \circ \delta_{c_{k-1}} \circ \dots \circ \delta_{c_1}$.

Since each $\delta_{c_i}$ is a permutation, and the composition of permutations is a permutation, the function $\hat{\delta}_w$ is also a permutation of $Q$ for any non-empty string $w$.

3. Every state is part of a cycle

A key property of any permutation $\pi$ on a finite set $S$ is that for any element $s \in S$, repeated application of $\pi$ eventually returns to $s$. That is, there exists an integer $k \ge 1$ such that $\pi^k(s) = s$. This is a direct consequence of the finiteness of the orbit of $s$.

Applying this to our DFA, let $q$ be any state in $Q$. Since we are given $\Sigma \ne \emptyset$, we can pick any symbol $c \in \Sigma$. From step 1, $\delta_c$ is a permutation. Thus, there exists an integer $k \ge 1$ such that $(\delta_c)^k(q) = q$. This is equivalent to $\hat{\delta}(q, c^k) = q$. Since $c^k$ is a non-empty string, this shows that every state $q \in Q$ is part of a cycle.

4. Proving "$L(M)$ is infinite or empty"

This statement is equivalent to proving: If $L(M)$ is non-empty, then $L(M)$ is infinite.

Assume $L(M)$ is non-empty. A language accepted by a DFA is infinite if and only if there exists a state that is reachable from the start state, can reach a final state, and is part of a cycle.

Let's check these conditions for the start state $q_0$:
1.  Reachability: $q_0$ is reachable from itself (by processing the empty string $\epsilon$).
2.  Ability to reach a final state: We assumed $L(M)$ is non-empty. By definition, this means there is at least one string $w \in L(M)$. For this string, we have $\hat{\delta}(q_0, w) \in F$. This shows there is a path from $q_0$ to a final state.
3.  Part of a cycle: From step 3, we proved that every state is part of a cycle, so $q_0$ must be part of a cycle. Let $v \ne \epsilon$ be a string such that $\hat{\delta}(q_0, v) = q_0$.

Since $q_0$ satisfies all three conditions, the language $L(M)$ must be infinite. Specifically, if $w \in L(M)$ and $\hat{\delta}(q_0, v)=q_0$, then all strings of the form $v^i w$ for $i \ge 0$ are in $L(M)$. Because $v$ is non-empty, this constitutes an infinite set of strings.

Conclusion

We have shown that if $L(M)$ is non-empty, it must be infinite. Therefore, $L(M)$ is either empty or infinite.

Because M′ is a DFA, δ′ must be total and single-valued. Hence for each c∈Σ, the map δ_c:Q→Q given by δ_c(q)=δ(q,c) is bijective (its inverse is δ′(·,c)). Thus M is a permutation (group) automaton: every letter acts as a permutation of Q.

Let G be the finite group generated by {δ_c : c∈Σ}, and let O be the orbit of q₀ under G: O={g(q₀) : g∈G}. Then
L(M) = { w∈Σ* : δ_w(q₀) ∈ F }.

If O∩F=∅, then L(M)=∅.

Otherwise pick s∈O∩F and u∈Σ* with δ_u(q₀)=s. Since Σ≠∅, choose any a∈Σ. The permutation δ_a has finite order m≥1, so δ_{a^m}=id. Hence for all k≥0,
δ_{u a^{mk}}(q₀)=δ_u(q₀)=s∈F,
so u a^{mk} ∈ L(M). These are infinitely many distinct words, so L(M) is infinite.

Therefore L(M) is either empty or infinite.