Suppose I have a normed vector space, $X$ and its dual $X'$. Then, in a loose sense, $X$ is "the same as" $X'$, as is written in my lecture notes when first introducing dual spaces. That is to say that $X$ is isometrically isomorphic to $X'$. Let us call this statement $(1)$.
Similarly, suppose that the second dual is such that $X''$ is "the same as" $X$; again, that they are isometrically isomorphic to each other. Let us call this statement $(2)$.
My question; does $(1) \implies (2)$ for any space that is isometrically isomorphic to its first dual? Or, could you have the first statement without the second necessarily following? Naturally we can take the dual of the dual of the ... dual ad infinitum; what I want to know is that when it is known that $X\sim X'$, does it automatically follow that $X\sim X''$, $X\sim X'''$, ... ?
*In this context, the dual space is the space of all continuous linear functionals between $X$ and the scalar field $\mathbb F$.
I am confused by OP's question but let me comment on it anyway. Saying that $X$ is the same as its dual is a stretch. This is only in the case of Hilbert spaces and some other more-or-lesss artificial examples.
For example, $\ell_1$ is separable, contrary to its dual $\ell_\infty$, so they are hardly the same. Actuall, the dual of $\ell_\infty$ has cardinality strictly greater than the cardinality of $\ell_\infty$ so the sitation is even worse. I would never call two sets of different cadinalities similar.
Update. Of course, by the Riesz representation theorem, every Hilbert space is isometric to its dual. One may coin other examples by taking a reflexive space $E$ and setting $X = E\oplus_2 E^\prime$. Using the James space one may get also non-reflexive examples but they are all artificial. Typically, $X$ is not isomorphic to $X^\prime$ and this is the case for all classical non-Hilbertian Banach spaces.