Q-1: I wasn't sure of the answer to your first question, so I did some searching around. Specifically, the group of $n \times n$ orthogonal matrices, denoted $O(n)$, does not appear to be a normal subgroup of the group of all invertible $n \times n$ matrices, denoted $GL_n(\mathbb{R})$.
What I mean by the above is that there exist orthogonal matrices $A$ and invertible matrices $S$ such that $SAS^{-1}$ is not orthogonal. Recall that the matrix of the linear tranformation given by $A$ under a different basis is given by $SAS^{-1}$ for some change-of-basis matrix $S$. Thus, the answer to your first question is no.
However, if we only use orthogonal change-of-basis matrices, i.e. we assume that the new basis is orthonormal, then the answer is yes. Observe, if $S$ is orthogonal, then so is $S^{-1}$, and we have
$$(SAS^{-1})^T(SAS^{-1}) = (S^{-1})^TA^TS^TSAS^{-1} = (S^{-1})^TA^TAS^{-1} = (S^{-1})^TS^{-1} = I$$
(we have just used that $A^TA = I$ for orthogonal matrices $A$). This shows that $SAS^{-1}$ is still orthogonal, so changing to a different orthonormal basis preserves orthogonality of the linear transformation.
Q-2: We do not have a concept of orthogonality, including orthogonal transformations, unless our vector spaces are indeed inner product spaces. As you point out, there is always a way to impose an inner product on a given vector space $V$, namely by picking a basis, using that basis to construct an isomorphism to $\mathbb{R}^n$, and then taking the dot product in $\mathbb{R}^n$. This is an example of something we usually call "non-canonical," which roughly means there were choices involved in the definition of this inner product. Namely, we had to choose a basis for $V$, and there are many different ways to do this, yielding many different inner products. Therefore, we do not typically use this inner product. Rather, we would hope that $V$ comes with a more "natural" or "canonical" inner product to define orthogonal transformations between arbitrary spaces.
Q-3: Again, if $V$ and $W$ are inner product spaces, only then may we discuss orthogonality, so let's assume they do indeed have inner product structures. Then as we found above, while a transformation may be orthogonal, its matrix with respect to a particular basis need not be. Once again, we will need to assume our bases $\mathcal{B}$ and $\mathcal{C}$ for $V$ and $W$ resp. are orthonormal with respect to the inner product structures on $V$ and $W$ resp. Under these circumstances, I believe we can conclude that the matrix for the orthogonal transformation $T$ is an orthogonal matrix (although I have not proven this fact. You should try to find a proof, or try to write a proof yourself!)
We have $\dim W$ since $(v_1, v_2, v_3)$ is linearly independent. To see that, suppose $$c_1v_1 + c_2v_2 + c_3v_3 = 0$$ for some $c_1,c_2,c_2\in\mathbb R$. This implies
\begin{align}
c_1 + c_2 + c_3 &= 0\tag 1\\
c_1 + c_3 &= 0\tag 2\\
c_1 + c_2 &= 0\tag 3\\
c_1 &= 0\tag 4
\end{align}
$(4)$ directly implies $c_1=0$, which in turn implies $c_2=c_3 = 0$ from $(2)$ and $(3)$.
To construct an orthogonal basis for $W$, there is a standard inductive algorithm called Gram-Schmidt. In general, suppose we have a basis $(v_1, \ldots, v_n)$ for a subspace $V$ of $\mathbb R^m$ where $n<m$ (the case where $n=m$ isn't interesting because we can just take the standard basis). The algorithm proceeds as follows:
- Let $V_1 = \operatorname{Span}(\{u_1\})$, where $u_1=v_1$.
- Given $V_i$, $i\leqslant 1\leqslant n-1$, set $V_{i+1} = V_i \cup \{u_{i+1}\}$ where $$u_{i+1} = v_{i+1} - P_{V_i}(v_{i+1}),$$
where $P_{\cdot}$ denotes the projection operator, i.e.
$$P_{V_i}(v_{i+1}) = \sum_{j=1}^i \frac{\langle v_{i+1}, v_j \rangle}{\langle v_j, v_j \rangle} v_j. $$
In this example, we have $u_1=v_1=(1,1,1,1)$, and then
\begin{align}
u_2 &= v_2 - P_{V_1}(v_2) = (1,0,1,0) - \frac{\langle (1,0,1,0), (1,1,1,1) \rangle}{\langle (1,1,1,1), (1,1,1,1) \rangle} \\
&= (1,0,1,0) - \frac24(1,1,1,1)\\
&= \left(\frac12,-\frac12,\frac12,-\frac12\right).
\end{align}
Finally,
\begin{align}
u_3 =&\ v_3 - P_{V_2}(v_2)\\
=&\ (1,1,0,0)\\ &- \left[\frac{\langle (1,1,0,0), (1,1,1,1) \rangle}{\langle (1,1,1,1), (1,1,1,1) \rangle}(1,1,1,1)+\right.\\ &\quad\quad\left.\frac{\langle (1,1,0,0), \left(\frac12,-\frac12,\frac12,-\frac12\right) \rangle}{\langle \left(\frac12,-\frac12,\frac12,-\frac12\right), \left(\frac12,-\frac12,\frac12,-\frac12\right) \rangle} \left(\frac12,-\frac12,\frac12,-\frac12\right) \right]\\
=&\ (1,1,0,0) - \left(\frac24 (1,1,1,1) + 0 \right)\\
=&\ \left(\frac12, \frac12, -\frac12, -\frac12\right).
\end{align}
(Notice that conveniently $v_3\perp u_2$, saving some computation). Hence, our orthogonal basis is
$$\left(u_1, u_2, u_3 \right) = \left(\left(1,1,1,1\right), \left(\frac12,-\frac12,\frac12,-\frac12\right),\left(\frac12, \frac12, -\frac12, -\frac12\right) \right)$$
To convert this orthogonal basis, we need only divide the basis elements by their norm, i.e. compute $e_i = \frac1{\|u_i\|}$ where $\|u_i\| = \langle u_i,u_i\rangle^{\frac12}$. I'll spare the computation, but the result is
$$(e_1, e_2, e_3) = \left(\left(\frac14, \frac14, \frac14, \frac14 \right),\left(\frac12,-\frac12,\frac12,-\frac12\right),\left(\frac12, \frac12, -\frac12, -\frac12\right) \right). $$
Best Answer
Let $P$ the change matrix from $V$ to $U$ i.e. we have
$$Pu_i=v_i,\;\forall i$$ We have
$$\delta_{i,j}=\langle Pu_i,Pu_j\rangle=(Pu_i)^T(Pu_j)=u_i^T(P^TP)u_j,\;\forall i,j$$ so we get
$$P^TP=I$$