Norm of a positive definite symmetric matrix by a vector $A$ is a positive definite symmetric matrix, thus it is possible to express $A$ as $A = Q\Lambda Q^{T}$ with $Q$ an orthogonal matrix and $\Lambda$ the eigenvalue matrix of $A$.
Apparently, it is possible to prove that $\|Ax\|_{2}= \|\Lambda x\|_{2}$.
I understand that the 2-norm remains unchanged when a vector is multiplied to the left by an orthogonal matrix.
Thus in our case: $\|Ax\|_{2}= \|Q\Lambda Q^{T}x\| = \|Q(\Lambda Q^{T}x)\| = \|\Lambda Q^{T}x\|$ because $Q$ is orthogonal. 
Now I know that I should be able to get rid of $Q^{T} ( \|\Lambda Q^{T}x\| = \|\Lambda x\| )$, but I don't know how to prove it, since this is not the case where an orthogonal matrix multiplies a vector to the left.
 A: Let $A = Q \Lambda Q^T$ and let $P \ne I$ be a permutation. Notice that
$$A = Q \Lambda Q^T = Q (PP^T) \Lambda (PP^T) Q^T = (QP) (P^T \Lambda P) (QP)^T.$$
Obviously, $QP$ is orthogonal and $P^T \Lambda P$ is a positive diagonal matrix (like $\Lambda$). Then, assuming that your statement is correct,
$$\|\Lambda x\| = \|Ax\| = \|(QP) (P^T \Lambda P) (QP)^T x \| = \|(P^T \Lambda P)x\|.$$
Taking $x = e_1$ (the first vector of the canonical base), we get that
$$\Lambda_1 = (P^T \Lambda P)_1 = \Lambda_i,$$
where $i$ depends on the permutation $P$. Doing this for all possible permutations $P$ gives us that $\Lambda_1 = \Lambda_i$ for all $i$, so $\Lambda = \lambda {\rm I}$ for some $\lambda > 0$. This means that your statement is correct only for
$$A = Q \Lambda Q^T = Q \lambda {\rm I} Q^T = \lambda Q Q^T = \lambda {\rm I}.$$
A: No, this is typically false.  For a $2$-by-$2$ counterexample, let $A=\begin{bmatrix}1&0\\0&2\end{bmatrix}$, $\Lambda=\begin{bmatrix}2&0\\0&1\end{bmatrix}$, and $Q=\begin{bmatrix}0&1\\1&0\end{bmatrix}$.  Then $\|Ax\|_2^2=x_1^2+4x_2^2$ and $\|\Lambda x\|_2^2 = 4x_1^2+x_2^2$. Thus the equality holds if and only if $x_1^2=x_2^2$, meaning it holds only on a union of two $1$-dimensional subspaces.
It is true however that the operator norms of $A$ and $\Lambda$ are equal, and your question might have arisen out of confusion with that.
If you want to have this be an identity that holds for all $x$, see Vedran's answer, which shows that it requires $A$ to be a scalar multiple of the identity. 
