Does eigenvalues of matrix change after multiplication by unitary matrix?

For a matrix $A \in \mathbb{C}^{n \times n}$, does multiplication by a unitary matrix $U$ change the eigenvalues of $A$? So for:

$$Ax = \lambda x \qquad \mathrm{and} \qquad AUy = \mu y $$

does $\lambda = \mu$ for some $x,y \in \mathbb{C}^n$?

I know the above is true for doing left and right multiplication by $U$:

$$ UAU^*y = \mu y \\ AU^*y = \mu U^* y \\ Az = \mu z \\ \therefore \mu = \lambda$$

(defining $z = U^* y$)

Under the guise that unitary matrices are simply rotations, it logically makes sense to me that $\mu$ and $\lambda$ should be identical, and only the the eigenvectors should be different. The statement is true for singular values (see here), but I'm having trouble proving it for eigenvalues (if it even is true).

Edit

After a quick example in python, I understand that the above is not true. So instead: where is my thought process going wrong with regard to how unitary matrices/rotations effect eigenvalues?

Edit 2

What I was really going for, but did not state correctly was that:

$$AX = \Lambda X \qquad \mathrm{and} \qquad AUY = MY \\$$

such that $ \lambda \in M, \forall \lambda \in \Lambda$, where $M$ and $\Lambda$ are diagonal matrices.

Your intuition is somewhat correct, but you are doing it wrong. When you mean a rotation, to what does it apply? To vectors, right?

So, if $A$ is the matrix of a linear map , we expect that if we rotates the axes, the the matrix changes, but the application is still the same, so the eigenvalues should be the same. Let's check this.

Let $f$ be a linear map from an $n$-dimensional vector space $E$ into itself, and $\mathcal B$ a basis of $E$. Let $A$ the matrix of $f$ in that basis. And let another basis $\mathcal B'$, with a matrix $P$ of change of basis, which is regular. That is, if $v$ is a vector in $E$ with coordinates $X\in\Bbb R^n$ in the basis $\mathcal B$, then the coordinates $X'$ in the basis $\mathcal B'$ satisfy (note the order):

$$X=PX'$$

Now, in the new basis, $f$ has another matrix representation $A'$, that is given by

$$A'=P^{-1}AP$$

How does it impact eigenvectors and eigenvalues? Let $u$ an eigenvector of $A$ for the eigenvalue $\lambda$, then $Au=\lambda u$ and, since $A=PA'P^{-1}$

$$PA'P^{-1}u=\lambda u$$

Hence

$$A'P^{-1}u=P^{-1}\lambda u=\lambda P^{-1}u$$

That is, $\lambda$ is also an eigenvalue of $A$, for the eigenvector $u'=P^{-1}u$. Which we can write $u=Pu'$. For the eigenvector $u'$, we simply apply the change of basis: it's the same vector in $E$, only its representation in a basis changes.

That is, the eigenvectors and eigenvalues of a linear map don't depend on the basis.

Note that for this we don't require $P$ to be unitary.