Is $\hat u \hat p$ also a projector if $\hat p$ is and if $\hat u$ is unitary? If $\hat p:\mathcal H\to\mathcal H_p\subseteq\mathcal H$ is a projection operator, and if $\hat u$ is unitary, then it's easy/trivial to show that $\hat u \hat p \hat u^{-1}$ is also a projector.
But if we interpret unitary $\hat u$ as a rotation in $\mathcal H$, then it seems (at least to me) intuitive that $\hat u\hat p$ should also be a projector, simply rotating that $\hat p:\mathcal H\to\mathcal H_p\subseteq\mathcal H$ subspace into another subspace $\hat u:\mathcal H_p\to\mathcal H_{up}$. But I can't seem to conjure up a proof that $\hat u\hat p$ is also a projector.
So, is it true to begin with, and if so, can you provide or cite such a proof? And if not, what's the problem with that "rotation reasoning"? (All I could think up myself is that maybe the rotated $\mathcal H_p$ subspace results in a subset of $\mathcal H$ that's no longer complete with respect to scalar multiplication or vector addition, or something like that. But unless I'm wrong, it remains complete wrt those operations, and hence remains a subspace.)
 A: No, it is not true.  Projectors are, by definition, idempotent; however, in general $(\hat u\hat p)(\hat u \hat p) \neq \hat u \hat p$.  As a trivial example, let $\hat u = -\mathbb I$, which maps $\psi\in\mathcal H \mapsto -\psi$.  Since $\hat u$ commutes with everything,
$$(\hat u \hat p)(\hat u \hat p)=\hat p^2 = \hat p  \neq \hat u \hat p$$
As a second counterexample, consider $\mathcal H = \mathbb R^3$ with $\hat p = \pmatrix{1&0&0\\0&1&0\\0&0&0}$ the projection onto the $(x,y)$ plane.  If $\hat u$ is a rotation about the $z$-axis, then it should be clear that $\hat u \hat p$ will not be idempotent (unless the rotation is trivial).  Furthermore, if $\hat u$ is e.g. a $\pi/4$ rotation about the $x$-axis, then successive applications of $(\hat u \hat p)$ to the vector $\psi =\pmatrix{0\\1\\0}$ will yield a sequence of vectors
$$(\hat u \hat p)^n \psi = \pmatrix{0\\1/\left(\sqrt{2}\right)^n\\0}$$
which is certainly not the kind of behavior we'd like in a projector.
A: J. Murray has given you a few counterexamples, so I'll address the other part of your question.
The problem with your intuition is that if $\hat p$ is an operator, $\hat U\hat p$ is not a rotation of $\hat p$. Consider a simple projection operator $\def\ket#1{\left|#1\right\rangle}\def\bra#1{\left\langle#1\right|}\def\ketbra#1{\ket{#1}\bra{#1}}\hat p=\ketbra{\psi}$ and a unitary operator $\hat U$ such that $\ket{\psi^\prime}=\hat U\ket{\psi}$.
The operator $\ketbra{\psi}$ projects a state onto $\ket{\psi}$. So if we interpret $\hat U$ as a rotation, then we should end up with an operator that projects a state onto $\ket{\psi^\prime}$. In other words, the operator rotated is given by:
$$\hat p^\prime=\ketbra{\psi^\prime}=\ketbra{\hat U\psi}=\hat U\ketbra{\psi}\hat U^\dagger=\hat U\hat p\hat U^\dagger$$
In other words, your intuition that just rotating a projector shouldn't change the fact that it is a projector is correct. It's just that rotating a projector is not done by $\hat U\hat p$ but instead by $\hat U\hat p\hat U^\dagger$.
