A projector equal to its own conjugate by a unitary For projector $p$, in finite dimension say, some unitaries $u, v$ does $upu^\dagger = vpv^\dagger$ implies $u = v$ ?
Intuitively, can we not say that a unitary is matrix permuting the basis and since $p$ is diagonal then obviously $u$ is $v$ ?
But for an exact proof ?
what if further, $p = upu^\dagger = vpv^\dagger$  ?
 A: The answer is no, your result does not follow.
To understand why, suppose we have some $N\times N$ projector $\mathbf{P}$, with a range space $\operatorname{Ran}(\mathbf{P})$ (the subspace that the projector projects into) of dimension $\dim(\operatorname{Ran}(\mathbf{P})) = n$ and kernel (nullspace)  $\ker(\mathbf{P})$ of dimension $\dim(\ker(\mathbf{P})) = m$ so that $N = n+m$. Intuitively, you can then do any nontrivial unitary transformation that unitarily transforms the range space alone, or any nontrivial unitary transformation that transforms the kernel alone, or any unitary transformation that is the product of the two, and the projector will have the same matrix.
To see this in detail, choose a basis $\{X_1, X_2, \cdots, X_n\}$ to span $\operatorname{Ran}(\mathbf{P})$ and basis $\{Y_1, Y_2, \cdots, Y_m\}$ to span $\ker(\mathbf{P})$. In this basis:
$$\mathbf{P} = \operatorname{diag}[\overbrace{1,\cdots,\,1}^{n\,\mathrm{terms}},\,\underbrace{0,\cdots,\,0}_{m\,\mathrm{terms}}] = \left(\begin{array}{cc}\mathbf{1}_{n\times n}&\mathbf{0}_{n\times m}\\\mathbf{0}_{m\times n}&\mathbf{0}_{m\times m}\end{array}\right)$$
where I have partitioned the matrix into the obvious blocks. Now consider any similarity transformation where we conjugate by a matrix of the form:
$$\left(\begin{array}{cc}\mathbf{U}_{n\times n} & \mathbf{0}_{n\times m}\\ \mathbf{0}_{m\times n} & \mathbf{U}_{m\times m}\end{array}\right)$$
where $\mathbf{U}_{n\times n}$ and $\mathbf{U}_{m\times m}$ are any unitary $n\times n$ and $m \times m$ matrices you can think of. We do the straightforward block-partitioned matrix calculation as follows:
$$\begin{array}{lcl}\left(\begin{array}{cc}\mathbf{U}_{n\times n} & \mathbf{0}_{n\times m}\\ \mathbf{0}_{m\times n} & \mathbf{U}_{m\times m}\end{array}\right)\, \mathbf{P}\, \left(\begin{array}{cc}\mathbf{U}_{n\times n}^\dagger & \mathbf{0}_{n\times m}\\ \mathbf{0}_{m\times n} & \mathbf{U}_{m\times m}^\dagger \end{array}\right) &= & \\
\left(\begin{array}{cc}\mathbf{U}_{n\times n} & \mathbf{0}_{n\times m}\\ \mathbf{0}_{m\times n} & \mathbf{U}_{m\times m}\end{array}\right)\, \left(\begin{array}{cc}\mathbf{1}_{n\times n}&\mathbf{0}_{n\times m}\\\mathbf{0}_{m\times n}&\mathbf{0}_{m\times m}\end{array}\right)\, \left(\begin{array}{cc}\mathbf{U}_{n\times n}^\dagger & \mathbf{0}_{n\times m}\\ \mathbf{0}_{m\times n} & \mathbf{U}_{m\times m}^\dagger \end{array}\right) &=& \\ \left(\begin{array}{cc}\mathbf{U}_{n\times n}\mathbf{1}_{n\times n}\mathbf{U}_{n\times n}^\dagger & \mathbf{0}_{n\times m}\\\mathbf{0}_{m\times n} & \mathbf{U}_{m\times m}\mathbf{0}_{m\times m}&\mathbf{U}_{m\times m}^\dagger \end{array}\right) &=& \left(\begin{array}{cc}\mathbf{1}_{n\times n}&\mathbf{0}_{n\times m}\\\mathbf{0}_{m\times n}&\mathbf{0}_{m\times m}\end{array}\right) = \mathbf{P}\end{array}$$ 
thus retrieving $\mathbf{P}$ and showing that there are huge classes of similarity transformations that will leave $\mathbf{P}$ invariant.
Even in the $2\times 2$ case, where $\operatorname{Ran}(\mathbf{P})$ and $\ker\mathbf{P})$ are one-dimensional (so there is no eigenvector "degeneracy" that can be used) we can still put $\mathbf{U}_{1\times1} = \exp(i\phi)$ for any real phase angle $\phi$.
