Orthogonal matrix norm If $H$ is an orthogonal matrix, then $||H||=1$ and $||HA||=||A||, \forall A$-matrix (such that we can writ $H \cdot A$).
What norm is this about?
 A: The operator norm
$$
\|A\|=\max\{\|Ax\|_2:\ \|x\|=1\},
$$
where $\|\cdot\|_2$ is the Euclidean norm, also satisfies those two equalities. They follow easily from the fact that $\|y\|_2^2=y^Ty$, so $$\|Hx\|_2^2=(Hx)^THx=x^TH^THx=x^Tx=\|x\|_2^2.$$
A: This holds for any norm induced by an inner product. This follows from
$$\|QA\|=\sqrt{(QA,QA)}=\sqrt{(Q^TQA,A)}=\sqrt{(A,A)} = \|A\|$$
With $Q$ an orthonormal matrix, i.e., $Q^{-1}=Q^T$.
A: The original question was asking about a matrix H and a matrix A, so presumably we are talking about the operator norm. The selected answer doesn't parse with the definitions of A and H stated by the OP -- if A is a matrix or more generally an operator, (A,A) is not defined (unless you have actually defined an inner product on the space of linear operators, but if that is the case it may be surprising to the OP)
The operator norm of AH would usually be defined by 
$$||HA||=\sup_{||x||=1} ||HAx||$$
where $||.||$ is any norm, such as the norm induced by the inner product (the euclidean norm in the case of the dot-product) .
$$=\sup_{||x||=1} \sqrt{(HAx,HAx)}$$
$$=\sup_{||x||=1} \sqrt{(H^*HAx,Ax)} $$
(definition of adjoint)
$$=\sup_{||x||=1} \sqrt{(Ax,Ax)} $$
 (H is an orthogonal matrix)
$$=\sup_{||x||=1} ||Ax||$$
(definition of operator norm)
$$=||A||$$
A: The frobenius norm, $||A||_F^2 = tr(A A^T)$ satisfies this property. You can see this by noting $|| H A||_F^2 = tr( HA (HA)^T) = tr(H A A^T H^T) = tr(H^T H A A^T) = tr( I A A^T ) = tr(A A^T) = ||A||_F^2$. 
