Eigenvalues of linear transformation of $ V\otimes V\otimes V\otimes V$ Let $V$ be a  $\mathbb{C}$-vector space.Let dim$ V = n$ and $W = V\otimes V\otimes V\otimes V$.
Let  $f:W \to W$ be a linear map defined by  $f(a\otimes b\otimes c\otimes d) = d\otimes a\otimes b\otimes c $ $(a,b,c,d\in V)$.
I want to prove that $f$ is diagnolizable and find all eigenvalues of $f$ and dimension of eigenspace corresponding to an eigenvalues.However, representation matrix of $f$ is too large to compute characteristic polynomial.Is there any good way?
 A: As the comment notes, is suffices to note that $f^4 = I$.  Thus, the minimal polynomial of $f$ divides $p(x) = x^4 - 1$, which factors into a product distinct linear factors over $\Bbb C$.  As such, $f$ must be diagonalizable with eigenvalues $\lambda \in \{\pm  1, \pm i\}$.
Now, the dimensions of the eigenspaces: because $f$ can be expressed with real coefficients, we must have $\dim E_{i} = \dim E_{-i}$.  We note that 
$$
E_i + E_{-i} = \ker(f^2 + I) = \{v \in \otimes^4 V: f^2(v) = -v\}
$$
Now, note that
$$
f^2([a \otimes b] \otimes [c \otimes d]) = [c \otimes d] \otimes [a \otimes b]
$$
Use this to conlcude that
$$
E_i + E_{-i} \cong (V \otimes V) \wedge (V \otimes V)
$$
From which we may deduce that $\dim(E_i + E_{-i}) = 2 \dim(E_i) = \binom{n^2}{2}$.  So, we have
$$
\dim E_i = \dim E_{-i} = \frac 12 \binom{n^2}2 = \frac{n^4 - n^2}{4}
$$
So, the combined dimension of the remaining eigenspaces is
$$
\dim(E_1 + E_{-1}) = n^4 - \dim(E_i + E_{-i}) = n^2 - \binom{n^2}2 = \binom{n^2 + 1}{2}
$$
It suffices now to determine $\dim(E_1)$, noting that
$$
E_1 = \{v : f(v) = v\}
$$ 
Take $\{e_1,\dots,e_n\}$ to be a basis of $V$.  We note that we can write any $v$ as
$$
v = \sum_{1 \leq i_1,i_2,i_3,i_4 \leq n}{\alpha_{(i_1,i_2,i_3,i_4)}}e_{i_1} \otimes e_{i_2} \otimes e_{i_3} \otimes e_{i_4}
$$
The condition that $f(v) = v$ is equivalent to the statement that
$$
\alpha_{(i_1,i_2,i_3,i_4)} = 
\alpha_{(i_4,i_1,i_2,i_3)} = 
\alpha_{(i_3,i_4,i_1,i_2)} = 
\alpha_{(i_2,i_3,i_4,i_1)}
$$
So, the dimension of $E_1$ is the number of such coefficients up to this symmetry.  That is, it is the number of distinct "necklaces" consisting of $4$ possibly repeating elements of $\{1,\dots,n\}$.

Using Burnside's lemma, we can compute
$$
\dim(E_1) = \frac {n^4 + n^2 + 2n}4
$$
With that we have 
$$
\dim(E_{-1}) = \binom{n^2 + 1}2 - \frac{n^4 + n^2 + 2n}4 = \frac{n^4 + n^2 - 2n}{4}
$$
