Is PCA invariant to orthogonal transformations? Let $A$ a be an $n$ x $p$ matrix and let $B$ be the transformed data set of $A$ under $Q$:
$$ B = A Q $$
where Q is a $p$ x $p$ orthogonal matrix:
$$ Q Q^T = I $$
$n$ is the number of samples (rows) and $p$ is the number of features (columns).
 
$A$ and $B$ are centred (zero mean for each feature).
Now suppose we perform PCA on both $A$ and $B$:
$$ \frac{1}{n-1} A^T A = V_A L_A V^T_A $$
$$ \frac{1}{n-1} B^T B = V_B L_B V^T_B $$
The question is, are the projections of $A$ and $B$ on their principal directions equal and why?
$$ A V_A = B V_B \ ? $$
That is, are the datasets identical upon "transforming" them with PCA?

Later edit: I simplified the question to only refer to orthogonal transformations (rotations). The original question referred to translations as well (the explanation in that case is trivial).
 A: The covariance matrix of $B$ is decomposed as:
$$ \frac{1}{n-1} B^T B = V_B L_B V^T_B $$
If we rewrite the covariance matrix of $B$ in terms of $A$ and $Q$ we have:
$$ \frac{1}{n-1} B^T B = \frac{1}{n-1} (AQ)^T AQ = \frac{1}{n-1} Q^T A^T A Q = Q^T (\frac{1}{n-1} A^T A) Q = Q^T V_A L_A V^T_A Q $$
We observe that the covariance matrices of $A$ and $B$ are generally different:
$$ V_A L_A V^T_A \neq Q^T V_A L_A V^T_A Q $$
We identify (wasn't sure about this step at first):
$$ V_B = Q^T V_A $$
$$ L_B = L_A $$
$$ V^T_B = (Q^T V_A)^T = V^T_A Q $$
And now we have:
$$ B V_B = A Q Q^T V_A = A V_A $$
Which means that the datasets are identical upon "transforming" them with PCA.

Many thanks to @amoeba and @Alex R.
A: Let $R$ be a rotation (i.e. an orthogonal matrix: $R^T=R^{-1}$). $A=U\Sigma V^*$ by SVD, and correspondingly $A^TA=V\Sigma^2 V^*$.
On the other hand if $B:=RA$, then:
$$B^TB=A^TR^TRA=A^TA=V\Sigma^2V^*.$$
So PCA is "invariant" under rotations applied to $A$ from the left. Specifically, this means that the covariance matrix is invariant. However the actual projections will differ. Specifically, $A=U\Sigma V^*$, whereas $B=RA=RU\Sigma V^*$, giving $B=W\Sigma V^*$, where $W:=R^*U$. 
