Does Preservation of induced Norm imply Unitarity? I was thinking about the definition of unitary and orthogonal matrices and how it relates to the inner product.
The definition of orthogonal/unitary Matrices is given by:
$O^T O = O O^T = \mathbb{1}$
$U^{\dagger}U = UU^{\dagger} = 1$
While the preservation of the induced Norm looks like this:
$\langle Ox\mid Ox\rangle = \langle x \mid x \rangle$
$\langle Ux\mid Ux\rangle = \langle x \mid x \rangle$
Now it is clear that by using the definition of matrix multiplication (or the inner product if you will) that you can let the left operator (or the right) in the second pair of equation jump out of its slot and act on the right slot instead. Then you immediately see that the first pair of equations is a sufficient criteria for satisfying the second pair of equations. My question is whether this also goes the other way around. Does the second pair of equations imply the first pair? 
Phrased differently: Does the second pair of equations imply the orthogonality/unitarity of the Matrices in question as defined above?
Edit: Yes, sorry for calling it a scalar product initially, it is a direct (wrong) translation of the german "Skalarprodukt"
And yes my title was also wrong, it's just the norm that is preserved.
 A: I will use "inner product" rather than "scalar product." (Edit: now fixed) The "scalar product" is often the name given to the map $K\times V\to V$ that is part of the definition of a vector space.
I will also note that I disagree with your definition of "preserves inner product". (Edit: Now fixed to “preserves norm”). In my opinion, it should be
$$\langle Ox,Oy\rangle = \langle x,y\rangle\qquad\text{for all }x,y\in V.$$
Instead, what you have is "preserves norm": since your equations amount to $\lVert Ox\rVert^2 = \lVert x\rVert^2$, and this yields $\lVert Ox\rVert = \lVert x\rVert$.
That said: I will treat both orthogonal and unitary together. I will use $U^*$ to denote the adjoint of $U$; the matrix is unitary if we are working over $\mathbb{C}$ and $U^*U = UU^* = I$, and is orthogonal if we are working over $\mathbb{R}$ and $U^*U=UU^*=I$. For matrices, the operation amounts to conjugate transpose over $\mathbb{C}$ and to transpose over $\mathbb{R}$.

Theorem. Let $V$ be a finite dimensional inner product space, and let $U$ be an operator on $V$. The following are equivalent:
  
  
*
  
*$UU^*=U^*U=I$.
  
*$\lVert Ux\rVert = \lVert x\rVert$ for all $x\in V$.
  
*$\langle Ux,Uy\rangle = \langle x,y\rangle$ for all $x,y\in V$.
  
*For any orthonormal basis $\beta$ of $V$, $U(\beta)= \{Ux\mid x\in\beta\}$ is an orthonormal basis of $V$.
  
*There exists an orthonormal basis $\gamma$ of $V$ such that $U(\gamma) = \{Ux\mid x\in \gamma\}$ is an orthonormal basis for $V$.
  

Proof. $1\implies 2$: $\lVert Ux\rVert^2 = \langle Ux,Ux\rangle = \langle x,U^*Ux\rangle = \langle x,x\rangle = \lVert x\rVert^2$, hence $\lVert Ux\rVert = \lVert x\rVert$.
$1\implies 3$: as above, $\langle Ux,Uy\rangle = \langle x,U^*Uy\rangle = \langle x,y\rangle$.
$2\implies 3$: Let $x$ and $y$ be arbitrary vectors. Then $\lVert A(x+y)\rVert^2 = \lVert x+y\rVert^2$. We therefore have:
$$\begin{align*}
\langle A(x+y),A(x+y)\rangle &= \langle Ax,Ax\rangle + \langle Ax,Ay\rangle + \langle Ay,Ax\rangle + \langle Ay,Ay\rangle\\
&= \lVert Ax\rVert^2 + \lVert Ay\rVert^2 + \langle Ax,Ay\rangle + \overline{\langle Ax,Ay\rangle}.\\
\langle x+y,x+y\rangle &=\langle x,x + \langle x,y
\rangle + \langle y,x\rangle + \langle y,y\rangle\\
&= \lVert x\rVert^2 + \lVert y\rVert^2 + \langle x,y\rangle + \overline{\langle x,y\rangle}.
\end{align*}$$
Now, since $\lVert Ax\rVert = \lVert x\rVert$ and $\lVert Ay\rVert = \lVert y\rVert$, we conclude that
$$\langle Ax,Ay\rangle + \overline{\langle Ax,Ay\rangle} = \langle x,y\rangle + \overline{\langle x,y\rangle}.$$
If we are working over the real numbers, this gives $2\langle Ax,Ay\rangle = 2\langle x,y\rangle$, and we conclude $\langle Ax,Ay\rangle = \langle x,y\rangle$, as desired.
If we are working over the complex numbers we conclude that $2\mathrm{Re}(\langle Ax,Ay\rangle) = 2\mathrm{Re}(\langle x,y\rangle)$, so the real parts are equal.
Now, replace $x$ with $ix$, we get
$$i\langle Ax,Ay\rangle + \overline{i\langle Ax,Ay\rangle} = i\langle x,y\rangle + \overline{i\langle x,y\rangle}$$
and therefore,
$$i\left(\langle Ax,Ay\rangle - \overline{\langle Ax,Ay\rangle}\right) = i\left(\langle x,y\rangle - \overline{\langle x,y\rangle}\right),$$
and hence $-2\mathrm{Im}(\langle Ax,Ay\rangle) = -2\mathrm{Im}(\langle x,y\rangle)$. Thus, the imaginary parts of $\langle Ax,Ay\rangle$ and $\langle x,y\rangle$ also agree, and we conclude that $\langle Ax,Ay\rangle = \langle x,y\rangle$, as desired.
$3\implies 2$: immediate.
$3\implies 1$: We consider the operator $I-U^*U$. Let $y\in V$. Then for any $x\in V$ we have
$$\begin{align*}
\langle x,(I-U^*U)y\rangle &= \langle x,y\rangle - \langle x,U^*Uy\rangle\\
&= \langle x,y\rangle - \langle Ux,Uy\rangle\\
&= \langle x,y\rangle - \langle x,y\rangle\\
&= 0.
\end{align*}$$
As this holds for all $x\in V$, it follows that $(I-U^*U)y = 0$. This holds for each $y\in V$, so $I=U^*U$. 
Now, this means that $U$ is one-to-one, and hence invertible; therefore, $U^*=U^{-1}$, so $UU^* = I$ as well. 
So now we know that 1, 2, and 3 are equivalent. This already answers your question.
$3\implies 4$: Note that 2 implies that $U$ is one-to-one, and since 3 implies 2, we know that $U$ is one-to-one. Thus, $U(\beta)$ has $|\beta|=\dim(V)$ elements. 
If $x,y\in\beta$, then $\langle x,y\rangle = \delta_{xy}$ (Kronecker's delta). Therefore, for $Ux,Uy\in U(\beta)$, $\langle Ux,Uy\rangle = \langle x,y\rangle = \delta_{xy} = \delta_{U(x),U(y)}$. Thus, $U(\beta)$ is an orthonormal subset of $V$ with $\dim(V)$ elements, and hence an orthonormal basis.
$4\implies 5$: immediate.
$5\implies 3$: Let $\gamma = [v_1,\ldots,v_n]$ and express $x$ and $y$ in terms of $\gamma$. Say
$$x = \sum_{i=1}^n \alpha_iv_i,\qquad y=\sum_{i=1}^{n}\beta_iv_i.$$
Then
$$\langle x,y\rangle = \left\langle \sum_{i=1}^n\alpha_iv_i ,\sum_{j=1}^n \beta_jv_j\right\rangle = \sum_{i=1}^n\alpha_i\overline{\beta_j},$$
where $\overline{\beta_j}$ is the complex conjugate of $\beta_j$ if $V$ is complex, and $\beta_j$ if $V$ is real.
On the other hand, we have
$$\begin{align*}
\langle Ux,Uy\rangle &= \left\langle \sum_{i=1}^n\alpha_iUv_i,\sum_{j=1}^n\beta_jUv_j\right\rangle\\
&= \sum_{i=1}^n\sum_{j=1}^n \alpha_i\overline{\beta_j}\langle Uv_i, Uv_j\rangle\\
&= \sum_{i=1}^n\sum_{j=1}^n \alpha_i\overline{\beta_j}\delta_{ij}\\
&= \sum_{i=1}^n \alpha_i\overline{\beta_i}\\
&= \langle x,y\rangle.
\end{align*}$$
The third equality holds because $U(\gamma)$ is an orthonormal set, so $\langle Uv_i,Uv_j\rangle = \delta_{ij}$.
This proves that 3, 4, and 5 are equivalent. As 3 was already known to be equivalent to 1 and 2, this proves the Theorem. $\Box$
