Answer: b) such sequence exists; c) such sequence doesn't exist.

a. Show the example of such a "consecutive" sequence which is different from $a_n = n$. Let $n = 2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\dots p_r^{\alpha_r}$ be the unique representation of $n$ as a product of primes, then the sequence $\{a_n\}$, $a_n = 2^{\alpha_2}3^{\alpha_1}5^{\alpha_3}\dots p_r^{\alpha_r}$ satisfies the statement of the problem.

Really, if $n = 2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\cdots p_r^{\alpha_r}p_{r+1}^0p_{r+2}^0\cdots$, $m = 2^{\beta_1}3^{\beta_2}5^{\beta_3}\cdots p_s^{\beta_s}p_{s+1}^0p_{s+2}^0\cdots$, $k = 2^{\gamma_1}3^{\gamma_2}5^{\gamma_3}\cdots p_t^{\gamma_t}p_{t+1}^0p_{t+2}^0\cdots$ and $a_n < a_m$, then $a_{kn} < a_{km}$, as
$$a_{kn} = 2^{\alpha_2+\gamma_2} 3^{\alpha_1+\gamma_1} 5^{\alpha_3+\gamma_3} \cdots p_{\max(r,t)}^{\alpha_{\max(r,t)}+\gamma_{\max(r,t)}} = a_k a_n,$$
$$a_{km} = 2^{\beta_2 + \gamma_2} 3^{\beta_1 + \gamma_1} 5^{\beta_3 + \gamma_3} \cdots p_{\max(s,t)}^{\beta_{\max(s,t)} + \gamma_{\max(s,t)}} = a_k a_m$$
and $a_{kn} = a_k a_n < a_k a_m = a_{km}$. The constructed sequence is non-trivial, because $a_2 = 3$, and is constructed using a permutation of the exponents of $n$ in its prime factorization by means of shifting the exponents of two and three.

b. Next, let
$$N(k) = \begin{cases} k-2, & \text{if } k = 4, 6, 8, \dots, \\ 1, & \text{if } k = 2, \\ k+2, & \text{if } k = 1, 3, 5, \dots \end{cases}$$
Then $N(k_1) \neq N(k_2)$ if $k_1 \neq k_2$. If now we construct a sequence in which for the number $n$ represented as a product of primes, we shift the exponents of $p_k$ to $p_{N(k)}$, then the constructed sequence satisfies the statement of the problem. Really, let for $n = 2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}7^{\alpha_4}\dots p_{r-1}^{\alpha_{r-1}}p_r^{\alpha_r}$
$$a_n = 5^{\alpha_1} 2^{\alpha_2} 11^{\alpha_3} 3^{\alpha_4} \cdots p_{N(r-1)}^{\alpha_{r-1}} p_{N(r)}^{\alpha_r}.$$
Then the sequence $\{a_n\}$ is "consecutive", because $a_{kn} = a_k a_n$ and $a_{kn} = a_k a_n < a_k a_m = a_{km}$.

Moreover, if $n \ge 2$ then $a_n \ne n$, because if $a_n = n$, then the following equalities hold: $\alpha_{N(k)} = \alpha_k$, $\alpha_{N(N(k))} = \alpha_{N(k)} = \alpha_k$, $\dots$, $\alpha_{N(...N(N(k)))} = \alpha_k$. But $N(...N(N(k)))$ is unrestrictedly large beginning with some iteration, so if there exists nonzero $\alpha_k$ in the prime factorization of $n$, then in its representation must exist nonzero powers of arbitrarily large prime numbers, which is a contradiction.

c. If for some sequence holds $a_1 \neq 1$, then there exists $l$ such that $a_l = 1$. Then $a_l < a_1$ and $a_{l2} < a_l = 1$ — contradiction, because $a_{l2} \in \mathbb{N}$.