I would like to know if there is some theorem that allows us to turn a process $(X_t,\mathcal{F}_t)_{t \geq 0}$ (under the canonical filtration) which is not a martingale into one, by maybe changing the filtration. In my particular case, I have some one dimensional process that has some trigonometric functions and a deterministic function of time.
More concretely, I have a process of the form $$X_t:=c(t) [\cos^{2/a} B_{f(t)} + \sin^{2/a} B_{f(t)}]$$
where $(B_t)_{t \geq 0}$ is a classical Brownian motion, and we can pick any $f(t)$ we like for our goal. I have checked that this is not a martingale using Ito's Lemma, for some functions $f(t)$ like $\log(t)$, $\sin(t), \cos(t)$, etc.
To the question in the title: this is not possible, since if $X$ is a martingale with respect to any filtration $\mathcal{G}$, then it is also a martingale with respect to its natural filtration. This is because if $X$ is a martingale with respect to $\mathcal{G}$, in particular $X$ must be adapted with respect to $\mathcal{G}$ and hence $\mathcal{F}_t \subseteq \mathcal{G}_t$ for every $t \geq 0$. Therefore, by the tower property of conditional expectations it holds for $s \leq t$ that
$$\mathbb{E}[X_t | \mathcal{F}_s] = \mathbb{E}[\mathbb{E}[X_t | \mathcal{G}_s] | \mathcal{F}_s]] = \mathbb{E}[X_s |\mathcal{F}_s] = X_s,$$ so $X$ is martingale with respect to $\mathcal{F}$. The answer to the question in the body is also negative, since the time-change of a martingale is again a (local) martingale, if we assume the time-change to be continuous. If we allow for general $f$, not necessarily increasing and continuous, the answer is also negative for your problem: Letting $a=1$ we have $\cos^2(B_{f(t)}) + \sin^2(B_{f(t)}) = 1$ and therefore $X_{t} = c(t)$ for any choice of $f$. Now if $c$ is not constant, no choice of $f$ will turn $X$ into a martingale.