Intuition behind Isomorphic spaces "Being the Same"

You are asking a good question.

Take this statement: For a field $F$, the following vector spaces are isomorphic:

• $F^{n^2}$
• The space $M_{n \times n}$ of $n \times n$ matrices over $F$
• The space $P_{n^2}$ of polynomials with degree less than $n^2$ with coefficients in $F$

The isomorphisms we are talking about in this example only concern the additive structure, the "+", and the scalar multiplication (multiplication with an element in $F$). If we are only allowed to do addition and multiplication with a scalar, then the two spaces behave exactly the same. But you are right that both spaces may allow us to do other things that you cannot naturally do in the respective other space.

But we can always define them in the other space! This is done in general as follows. Take your isomorphism $\phi$, for example $\phi \colon M_{n \times n} \to P_{n^2}$. In $P_{n^2}$ we have differentiation, given by a map $D \colon P_{n^2} \to P_{n^2}$. How can we define differentiation in our matrix space? There is only one way if we want our new definition to be isomorphic to the definition on $P_{n^2}$. We have to define our new differentiation on matrices as $D_M := \phi^{-1} \circ D \circ \phi$. In other words: $$ D_M \colon M_{n \times n} \to M_{n \times n} \\ m \mapsto \phi^{-1}(D(\phi(m))) $$

For example, let's "differentiate" the matrix $$ \pmatrix{ 1 & 2 \\ 3 & 4 } $$

As a polynomial, this is $f(x) = x^3 + 2 x^2 + 3x + 4$ (depends on your choice of $\phi$!). So the derivative is $f'(x) = 0x^3 + 3x^2 + 4x + 3$. As a matrix, this is $$ \pmatrix{ 0 & 3 \\ 4 & 3 } $$

This is your "derivative" of the matrix.

Because "isomorphic" literally means "same structure", so thats because two isomorphic spaces are treated as the same. If you think about it, an isomorphism is a bijection with special conditions between the operations in the two different spaces. This basically means that if two spaces are isomorphic, their structure will be the same because the operations work in the same way. In other words, two isomorphic spaces are two different representations of the same structure.

Let $V$ be the set of real polynomials of degree at most $1$. Then $V$ is isomorphic to $\mathbb R^2$ under the isomorphism $\phi:ax+b \mapsto (a,b)$.

Does this imply that the elements of $V$ are the same as the elements of $\mathbb R^2$?

Clearly not: $V$ contains functions, $\mathbb R^2$ contains points.

Does this imply that the elements of $V$ behave exactly in the same way as the elements of $\mathbb R^2$?

Yes, their linear properties are the same in the sense that each linear operation in $V$ is mirrored in $\mathbb R^2$ via $\phi$. But not all properties are mirrored: Every non-constant polynomial of degree $1$ has a real zero. This sentence doesn't even make sense in $\mathbb R^2$. But then this sentence is not about linear properties of functions.