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Is it true that the all the functions which have the same $L^{2p}$ norms must be the same, up to isometries? More precisely I would like to show that given $f_1,f_2 : \mathbb R \to \mathbb R$ smooth as needed, $1$-periodic, and s.t. $$ \int_0^1 f_1^{2p} = \int_0^1 f_2^{2p}, \quad \forall p \in \mathbb N $$ then there is $t \in \mathbb R$ s.t. $f_1(x) = f_2(x+t)$ or $f_1(x) = f_2(t-x)$, $\forall x \in [0,1]$.

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    That doesn't quite work. Let $f_1$ consist of two bumps, and $f_2$ have on bump the same, but the other bump slightly translated. – 2017-01-12
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    You can also play games with signs since you always have the power $2p$. Indeed, $f_1 = f$ and $f_2 =-f$ will have your stated property but are not translates of each other in general. You can generalize this in the style of the example given by @DanielFischer as well. Indeed, take $f_1$ with two bumps of disjoint support, and then take $f_2$ to be the same bumps, but one of them with the sign flipped. – 2017-01-12
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    @Glitch the OP is asking whether they are the same up to an *isometry* (which includes rotations), so your example doesn't work. – 2017-01-12
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    @OpenBall Fair enough. I was addressing the "more precisely" part, where OP restricts to translation. – 2017-01-12
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    @Glitch oh, I didn't read that well (I thought that the rest was just a rephrasing of the question) – 2017-01-12

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As commenters pointed out, this is not true. To clarify the picture, introduce the distribution function $$\lambda_f(t) = \mu(\{x : |f(x)|\ge t\}),\qquad t>0$$ One can show that $$ \|f\|_p^p = \int_0^\infty t^{p-1} \lambda_f(t)\,dt $$ Hence, the equality $\lambda_f = \lambda_g$ implies $\|f\|_p=\|g\|_p$ for all $p\in [1,\infty)$ (actually, for $p=\infty$ too).

To visualize what can be done with $f$ while preserving its distribution function, think of sliding the level sets of $f$ to the left or right, ensuring they remain properly stacked. For example, if the graph of $f$ is a triangle, it can be deformed to any other triangle of the same base and height.