Show that if the matrix of $T$ with respect to all bases of $V$ is same, then $T = alpha I$ where $T$ is a linear operator on $V$

To complement the other answers, here is a straightforward method, that just uses the definition of the matrix of $T$.

Fix a basis $v_1,ldots,v_n$. We have begin{align} Tv_1&=T_{11}v_1+T_{21}v_2+ldots+T_{n1}v_n\ Tv_2&=T_{12}v_1+T_{22}v_2+ldots+T_{n2}v_n\ end{align} As the matrix of $T$ is the same with respect to the basis $v_2,v_1,v_3,ldots,v_n$, begin{align} Tv_2&=T_{11}v_2+T_{21}v_1+ldots+T_{n1}v_n\ Tv_1&=T_{12}v_2+T_{22}v_1+ldots+T_{n2}v_n\ end{align} Comparing the expressions, and using the linear independence of the basis, we get $T_{11}=T_{22}$ and $T_{21}=T_{12}$. As we can do this with any pair of elements in the basis we get $$ T_{11}=T_{22}=cdots=T_{nn} $$ and $T_{kj}=T_{jk}$ for any $kne j$. Now consider the basis $$ v_1-v_2,v_1+v_2,v_3,ldots,v_n. $$ Using this basis, begin{align} Tv_1-Tv_2&=T_{11}(v_1-v_2)+T_{21}(v_1+v_2)+T_{31}v_3+cdots+T_{n1}v_n\ Tv_1+Tv_2&=T_{12}(v_1-v_2)+T_{22}(v_1+v_2)+T_{32}v_3+cdots+T_{n2}v_n\ end{align} Adding, $$ 2Tv_1=(T_{11}+T_{12}+T_{21}+T_{22})v_1+(T_{21}+T_{22}-T_{11}-T_{12}),v_2+(T_{31}+T_{32})v_3+cdots $$ The previous relations we found allow us to reduce this to $$ Tv_1=(T_{11}+T_{12})v_1+0,v_2+T_{32}v_3+cdots $$ Comparing with the first $Tv_1$ we obtain $T_{11}+T_{12}=T_{11}$, so $T_{12}=0$. As we can do this for any pair $kne j$ of indices, we obtain $T_{kj}=0$. Then $T=alpha,I$, with $alpha=T_{11}$.

There are several possible proofs. One goes like this - it uses elementary matrices, but implicitly.

Suppose first of all that there is a vector $v$ such that $v, T v$ are linearly independent. Choose a basis that starts with $v, T v$. Then the first column of $T$ will have a $1$ in the second position, and a zero elsewhere. Now choose a basis that starts with $v, v + T v$. This time the first column of $T$ will have two $1$'s in the first two positions, and then all zeroes.

Since this is a contradiction, we will have that $v, T v$ are always dependent, for all vectors $v$. Let $v, w$ be two independent vectors (if there aren't any, the space has dimension $1$, and we are done). Let $T v = a v$, $T w = b v$, and $T (v + w) = c (v + w)$ for some $a, b, c$. Then $$ c v + c w = T(v + w) = T v + T w = a v + b w, $$ so that $a = c = b$, and you are done.

It seems that this is related to the isotropic matrices, i.e. matrices proportional to the identity. When you use a change of basis you transform $T$ into something else (in general components of the transformed matrix are different, but this does not occur in here). Recall the transformation of $T$ into $Q$: $Q=V^{-1}T : V$. You can replace $T$ by $alpha I$ to check the result for any $V$ (hint: pull out the scalar and use properties of the identity).

$B^{-1}AB=A$ is the same as $AB=BA$ for every invertible $B$. Now use elementary non-diagonal matrices $B$ to prove the result.