Do eigenvalues depend on the choice of basis?

Recall the definition:

Let $f$ be an endomorphism of a vector space $V$, then $\lambda$ is an eigenvalue of $f$ if there exists some non-zero $v \in V$ such that $f(v)=\lambda v$.

This does not involve a basis of the space at all. Thus it must be invariant under change of basis.

No, eigenvalues are invariant to the change of basis, only the representation of the eigenvectors by the vector coordinates in the new basis changes.

Indeed suppose that

$$Ax=\lambda x$$

and let consider the change of basis $x=My$ then

$$Ax=\lambda x\implies AMy=\lambda My\implies M^{-1}AMy=\lambda y \implies By=\lambda y$$

The whole point of eigenvalues and eigenvectors is to produce a bunch of axes that define your skewy transformation, so that your skewy transformation becomes a scaling transformation on these axes. If anything, this gives you a nice basis (one in which your matrix is diagonal, i.e. scaling). Your eigenvalues are clearly the same in the eigenbasis as in any other basis (they're across the diagonal), so the eigenvalues are the same in all bases.