The characteristic polynomial of a rotation satisfies the relation $f(\lambda) = (-\lambda)^n f(\lambda^{-1})$ In the book of Linear Algebra by Greub, at page 237, it is asked to prove that

The characteristic polynomial of a proper rotation satisfies the relation
  $f(\lambda) = (-\lambda)^n f(\lambda^{-1})$

where a rotation defined as an isometry on a finite dimensional real inner product  space $E$.
I have tried the following:
$$(-\lambda)^n det(\phi - \lambda^{-1}i) = det(-\lambda \phi + i),$$
and we know that $det\phi = \pm 1$, so in a sense the above equation is similar to $\det(\phi - \lambda i)$, however the entries of the matrix of $\phi$ and $i$ are obviously different, and applying the definition of determinant given in the book, I couldn't arrive anywhere, so I would appreciate any help or hint.
 A: The problem as originally stated by our OP onurcanbektos can't be quite right, as is illustrated by the following example which is based upon the identiy map $I$ on $\Bbb R^3$:  consider $\phi = -I$ on $\Bbb R^3$:  the characteritic polynomial of $\phi$ is
$f(\lambda) = \det (-I - \lambda I) = \det((1 + \lambda)(-I)) = (1 + \lambda)^3 \det(-I) = -(1 + \lambda)^3; \tag 1$
then
$f(\lambda^{-1}) = -(1 + \lambda^{-1})^3, \tag 2$
and
$(-\lambda)^3 f(\lambda^{-1}) = -\lambda^3 (-(1 + \lambda^{-1})^3) = (1 + \lambda)^3; \tag 3$
here we have an example that
$f(\lambda) = -(-\lambda)^3 f(\lambda^{-1}); \tag 4$
now if our isometry is $I$, we find
$f(\lambda) = \det (I - \lambda I) = \det((1 - \lambda)I) = (1 - \lambda)^3 \det (I) = (1 - \lambda)^3, \tag 5$
and
$f(\lambda^{-1}) = (1 - \lambda^{-1})^3, \tag 6$
so that
$(-\lambda)^3f(\lambda^{-1}) = (-\lambda)^3 (1 - \lambda^{-1})^3 = -(\lambda(1 - \lambda^{-1}))^3$
$= -(\lambda - 1)^3 = (1 - \lambda)^3 = f(\lambda). \tag 7$
If we compare (4) and (7) we might speculate that there is a factor or other information related to $\det \phi$ which is missing from the equation as stated in the text of the question,
$f(\lambda) = (-\lambda)^n f(\lambda^{-1}); \tag 8$
in fact, I researched Greub's book, Linear Algebra, and discovered that the actual problem quoted here, problem 9 on p. 237, reads
$"$9.)  Prove that the characteristic polynomial of a proper rotation satisfies 
$f(\lambda) = (-\lambda)^n f(\lambda^{-1})." \text{[italics mine.]} \tag 9$
Greub defines $\phi$ to be a proper rotation if it is an isometry with $\det \phi = 1$.
Based on this understanding, problem (9), p. 237 of Greub may be solved as follows:
$\phi$ is an isometry of the real inner product space $E$ if and only if it preserves inner products; that is, for $x, y \in E$ we have
$\langle \phi x, \phi y \rangle = \langle x, y \rangle; \tag {10}$
this of course implies
$\langle x, \phi^T \phi y \rangle = \langle x, y \rangle, \tag{11}$
and also
$\langle \phi^T \phi x, y \rangle = \langle x, y \rangle; \tag{12}$
as is well-known, (11) and (12) imply
$\phi^T \phi = I = \phi \phi^T, \tag{13}$
which shows that $\phi$ is invertible and that
$\phi^{-1} = \phi^T; \tag{14}$
such transformations are also called orthogonal.  Since
$\det \phi^T = \det \phi, \tag{15}$
a well-known property of determinants, it follow from (13) that
$(\det \phi)^2 = (\det \phi^T)(\det \phi) = \det \phi^T \phi = \det I = 1; \tag{16}$
that is,
$\det \phi = \pm 1; \tag{17}$
we also have
$\det \phi^{-1} \det \phi = \det \phi^{-1} \phi = \det I = 1; \tag{18}$
combining this with (15) and (17) we find
$\det \phi^T = \det \phi^{-1} = \det \phi = \pm 1; \tag{19}$
if $\phi$ is proper, we choose the $+$ sign in (19).
With these preliminaries completed, we thus have:
$(\phi - \lambda I)^T = \phi^T - \lambda I, \tag{20}$
whence
$f(\lambda) = \det(\phi - \lambda I) = \det ((\phi - \lambda I)^T) = \det (\phi^T - \lambda I); \tag{21}$
then from (14),
$f(\lambda) = \det (\phi^T - \lambda I) = \det (\phi^{-1} - \lambda I)$
$= \det (\phi^{-1}(I - \lambda \phi)) = \det \phi^{-1} \det(I - \lambda \phi) =  \det(I - \lambda \phi), \tag{22}$
where we have exploited the $+1$ case of (19); we thus conclude:
$f(\lambda) = \det(I - \lambda  \phi) = \det (-\lambda I)\det(\phi - \lambda^{-1}I) = (-\lambda)^nf(\lambda^{-1}), \tag{23}$
establishing (8) for proper rotations.  
In the event that $\phi$ is not a proper rotation, i.e., that $\det \phi = -1$, we can go back to (22) and pick up the derivation to find
$f(\lambda) = \det (\phi^T - \lambda I) = \det (\phi^{-1} - \lambda I)$
$= \det (\phi^{-1}(I - \lambda \phi)) = \det \phi^{-1} \det(I - \lambda \phi) =  -\det(I - \lambda \phi), \tag{24}$
from which it follows as in (23),
$f(\lambda) = -(-\lambda)^n f(\lambda^{-1}); \tag{25}$
we note these formulas are consisent with (4) and (8).
