Consider a matrix depending on a parameter $\xi$: $$ A(\xi)=\begin{pmatrix} a_{11}(\xi) & a_{12}(\xi) & \cdots & a_{1n}(\xi) \\ a_{21}(\xi) & a_{22}(\xi) & \cdots & a_{2n}(\xi) \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1}(\xi) & a_{n2}(\xi) & \cdots & a_{nn}(\xi)\end{pmatrix}, $$ where $\xi \in \mathbb{R}.$ We assume all the entries $a_{ij}(\xi)$ is homogenous with degree $1,$ that is $$ a_{ij}(k \xi)=k a_{ij}(\xi), \qquad \forall k>0. $$ I am not sure whether or not the eigenvalues of $A(\xi)$ are also homogenous with degree $1$?
Inspired by @NicNic8 's answer, now we make an additional assumption(maybe could be removed) that "all the eigenvalues of $A(\xi)$ are distinct for any $\xi \in \mathbb{R}$."
The characteristic polynomial of $A(\xi)$ is \begin{equation} \lambda^n(\xi)+c_1(\xi)\lambda^{n-1}(\xi)+\cdots+c_n(\xi)=0, \end{equation} and clearly $c_i(\xi)$ is a homogenous function with degree $i.$ Then for any positive constant $k,$ we have $$ \lambda^n(k\xi)+kc_1(\xi)\lambda^{n-1}(k\xi)+\cdots+k^nc_n(\xi)=0, $$ or equivalently $$ (\lambda(k\xi)/k)^n+c_1(\xi)(\lambda(k\xi)/k)^{n-1}+\cdots+c_n(\xi)=0. $$ We denote the eigenvalues of $A(\xi)$ by $\{\lambda_1(\xi),\cdots,\lambda_n(\xi)\}$ and denote the eigenvalues of $A(k\xi)$ by $\{\lambda_1(k\xi),\cdots,\lambda_n(k\xi)\},$ then we have the following two sets are equal for any $\xi \in \mathbb{R}$:
\begin{equation} \{\lambda_1(\xi),\cdots,\lambda_n(\xi)\}=\{\lambda_1(k\xi)/k,\cdots,\lambda_n(k\xi)/k\}, \quad \forall k>0. \end{equation} since the roots of the reciprocally third equation and the last equation are the same. That is the conclusion of @lonzaleggiera .
However, we still need to show that $$ \lambda_i(\xi)=\lambda_i(k\xi)/k, \quad \text{for} \ i=1,\cdots,n, \forall \xi \in \mathbb{R}, \forall k>0. $$ For example, how to exclude the following situation $$ \lambda_1(\xi_0)=\lambda_2(k\xi_0)/k_0, \quad \text{and} \quad \lambda_1(\xi_0)\neq\lambda_1(k\xi_0)/k_0 $$ at some $(\xi_0,k_0)$?
$\def\ed{\stackrel{\text{def}}{=}}$ If $\ a_{ij}(\xi)\ $ are all homogeneous with degree $\ r\ $, then $\ A(\xi)=\xi^rA(1)\ $. Let $\ \mu_1,\mu_2,\dots,\mu_m\ $ be the eigenvalues of $\ A(1)\ $ and $\ L_1,L_2,\dots,L_m\ $ the corresponding eigenspaces. Then for $\ v\in L_i\ $ \begin{align} A(\xi)v&=\xi^r A(1)v\\ &=\xi^r\mu_iv\ . \end{align} Thus, for each $\ i\ $, $\ L_i\ $ is an eigenspace of $\ A(\xi)\ $ corresponding to an eigenvalue $\ \xi^r\mu_i\ $. Conversely, if $\ \lambda_i(\xi)\ $ is an eigenvalue of $\ A(\xi)\ $ corresponding to an eigenvector $\ v\ ,$ then \begin{align} \lambda_i(\xi) v&=A(\xi)v\\ &=\xi^rA(1)v\ . \end{align} Therefore, for $\ \xi\ne0\ $, $\ v\ $ must be an eigenvector of $\ A(1)\ $ corresponding to an eigenvalue $\ \frac{\lambda_i(\xi)}{\xi^r}\ $, and must belong to $\ L_j\ $ for some $\ j\ ,$ thus giving $\ \lambda_i(\xi)=\mu_j\xi^r\ $. Since $\ A(0)=0\ $ has just the single eigenvalue $\ 0\ $, it remains true that $\ \mu_j\xi^r\ $ is an eigenvalue of $\ A(\xi)\ $ for $\ \xi=0\ $. Thus, for all $\ \xi\ $, the eigenvalues of $\ A(\xi)\ $ are $$ \mu_1\xi^r,\mu_ 2\xi^r,\dots,\mu_m\xi^r\ . $$ Note that this only shows that there exists a set of homogeneous functions $\ \lambda_i(\xi)\ed\mu_{p(i)}\xi^r\ $, for some $\ p:\{1,2,\dots,n\}\rightarrow \{1,2,\dots,m\}\ $, such that $\ \lambda_1(\xi),\lambda_2(\xi),\dots,\lambda_n(\xi)\ $ are the eigenvalues of $\ A(\xi)\ $. Since the indexing of the eigenvalues of $\ A(\xi)\ $ can be arbitrarily chosen, however, the functions $\ \lambda_1(\xi),\lambda_2(\xi),\dots,\lambda_n(\xi)\ $ are not uniquely defined, and it will always be possible to choose them to be non-homogeneous.
Reply to OP's comment below
It's unnecessary to assume anything about the multiplicities (whether geometric or algebraic) of the eigenvalues of $\ A(\xi)\ $, since it follows automatically from the fact that $\ A(\xi)=$$\,\xi^rA(1)\ $ that for $\ \xi\ne0\ $ they must be identical to those of the eigenvalues of $\ A(1)\ $. If $\ m_i\ $ is the algebraic multiplicity of $\ \mu_i\ $, then the characteristic polynomial of $\ A(1)\ $ is $$ \prod_{i=1}^m\big(x-\mu_i\big)^{m_i}\ , $$ and from the Cayley-Hamilton theorem, we get $$ 0_{n\times n}=\prod_{i=1}^m\big(A(1)-\mu_i\big)^{m_i}\ . $$ Multiplying this equation by $\ \xi^{rn}\ $, and noting that $\ \sum_\limits{i=1}^mm_i=n\ $, we get \begin{align} 0_{n\times n}&=\prod_{i=1}^m\big(\xi^rA(1)-\xi^r\lambda_i\big)^{m_i}\\ &=\prod_{i=1}^m\big(A(\xi)-\xi^r\lambda_i\big)^{m_i}\ . \end{align} Thus, the characteristic polynomial of $\ A(\xi)\ $ must be $$ \prod_{i=1}^m\big(x-\xi^r\mu_i\big)^{m_i}\ , $$ and $\ m_i\ $ is the algebraic multiplicity of the eigenvalue $\ \xi^r\mu_i\ $.