Are isometric normed linear spaces isomorphic?

This is true for real vector spaces by the Mazur-Ulam theorem which states that a surjective distance-preserving linear map of one real normed space onto another is an affine map. Indeed, if $f: X \to Y$ is such a map then $g(x) = f(x) - f(0)$ is a linear and onto isometry. The inverse of $g$ is of course linear and isometry, too, so in particular, $X$ and $Y$ are linearly isomorphic.

For a proof of this, one can't do better than refer to Väisälä's recent note which appeared in the Monthly, see here for the paywalled published version. References to the original works are given there.

Further remarks:

1. One corollary of the Mazur-Ulam theorem is that the group of isometric and onto self-maps of a real normed space is isomorphic to $O(X) \ltimes X$, where $O(X)$ denotes the group of linear isometries and acts on $X$ in the obvious way. This generalizes the usual description of Euclidean isometry group $O(n) \ltimes \mathbb{R}^n$ nicely.

2. If surjectivity of $f$ is dropped, the conclusion that $f$ needs to be affine is wrong. The example given by Väisälä is mentioned also this thread here: if we equip $\mathbb{R}^2$ with the max-norm and take $\phi:\mathbb{R} \to \mathbb{R}$ to be a non-linear $1$-Lipschitz map then the map $f: \mathbb{R} \to \mathbb{R}^2$ given by $f(x) = (x,\phi(x))$ is isometric but not affine.

3. Isomorphic Banach spaces need not be isometric, as you showed yourself in this recent thread.

4. As far as I know, the case of complex Banach spaces doesn't have such a nice formulation, since we need to ensure complex linearity in order to avoid silly counterexamples that stem from complex conjugation.

5. Finally, the Mazur-Ulam theorem was mentioned in this MO thread, where Bill Johnson mentions a nice generalization by Figiel which shows in some sense that the counterexample given in 2. is worst possible.

6. Tangentially relevant is also this MO thread on notions of isomorphisms of Banach spaces.