Eigen values of a transpose operator Let $T$  be linear operator on $M_{nxn}(R)$ defined by $T(A)=A^t.$ 
Then $\pm 1$ is the only eigen value.
My try : 
Let $n=2,$ then $[T]_{\beta}$ = $ 
\begin{pmatrix}
  a & 0 & 0 & 0 \\
  0 & c & 0 & 0 \\
  0 & 0 & b & 0 \\
  0 & 0 & 0 & d 
\end{pmatrix}$
Then the corresponding eigen values are $\lambda = a,b,c,d  $
How come i can claim it's eigen value is $\pm 1 ?$

What you are trying to show is that if there is an $n\times n$ real matrix $A$ (not equal to the zero matrix) such that $T(A)=\lambda A$ for some scalar $\lambda,$ then $\lambda=1$ or $\lambda=-1.$  Moreover, you are trying to show that $T$ has eigenvalues of $\pm 1,$ meaning that you need to exhibit some particular matrices $A_1,A_2$ of the kind I mentioned above, such that $T(A_1)=A_1$ and $T(A_2)=-A_2.$ (The matrix in your question gives you a start on finding such a matrix $A_1$. I leave it to you to find such an $A_2.$)
To prove these are the only ones, suppose $T(A)=\lambda A$ and consider $$T\bigl(T(A)\bigr).$$
