Can we find a stochastic matrix $Q$ such that $M = \frac{1}{C} \sum_{i=1}^C Q^i$ where $M$ is a stochastic matrix? Given a stochastic matrix $M$, can we always find another stochastic matrix $Q$ such that $M$ can be written by the mean of a geometric sum $M = \frac{1}{C} \sum_{i=1}^C Q^i$ for any $C > 1$?
My definition of stochastic matrix is a real square matrix whose components are non negative, with each row summing to 1.
The intuition is from the image of $f(x)=\frac{1}{C} (\frac{1-x^{C+1}}{1-x}-1)= \frac{1}{C} \sum_{i=1}^C x^i$. $f$ is strictly monotone in $[0,1)$ and $f\left([0,1)\right)= [0,1)$. We can extend $f$ to have a bijection from $[0,1]$ to $[0,1]$. As any eigenvalue of a stochastic matrix is $\le 1$, can we quantify the image of $g(Q) = \frac{1}{C} \sum_{i=1}^C Q^i$ on the space of stochastic matrix? Can that image cover the whole space of stochastic matrix (which could be seen as a product of simplices)?
Extra conditions: what if the Markov chain associated to $M$ is irreducible (and positively recurrent because the state space is finite)? even ergodic?
If we put stronger condition to make $M=P^{-1} \Delta P$ diagonalizable (for example, if reversible), we can have solution $D$ for eigenvalues: $g(D) = \Delta$. However is $Q=P^{-1} D P$ still stochastic?
 A: The question is stated in full generality for a stochastic matrix $M$, and a fixed $C$.
For a negative answer it is thus enough to consider a simple, very specific $M$. (But $C$ is still "general enough" and fixed for the issue.)
I decided to work with the example of a double stochastic matrix
$$
M = \begin{bmatrix}p & q\\ q & p\end{bmatrix}\ ,\ 0<p,q<1\ ,\ p+q=1\ .
$$
This is the case of a symmetric, diagonalizable $M$, its eigenvalues are $1=p+q$ and $p-q$.
Assume there is a matrix $Q$ with the property
$$
M = \frac 1C(Q+Q^2+\dots+Q^C)=g_C(Q)=g(Q)\ .
$$
(The notation $g$ is used for short for $g_C$, if $C$ is clear in the context.)
Then $M$ and $Q$ are commuting, simultaneously diagonalizable, let $P$ be the base change matrix as in the OP, and if $s,t$ are the eigenvalues of $Q$ (in the one good order matching via $g$ the order of $1$, $p-q$)...
$$
\begin{aligned}
M &=
P^{-1}
\begin{bmatrix}1 &\\ &p-q\end{bmatrix}
P\ ,\\
Q &=
P^{-1}
\begin{bmatrix}s & \\ & t\end{bmatrix}
P\ ,\\
P &=
\begin{bmatrix}1 & 1\\ 1 & -1\end{bmatrix}
\ ,\\[3mm]
1 &= \frac 1C(s+s^2+\dots+s^C)=g(s)\ ,\text{ so $s=1$,}\\
p-q 
&= \frac 1C(t+t^2+\dots+t^C)=g(t)\ .
\end{aligned}
$$
Now consider the case $C=2$, $g(t)=g_2(t)=\frac 12(t+t^2)$, $g:(-1,1)\to[-1/8,1)\subset(-1,1)$. If $p-q$ is in the image of $g$, then any $t$ in the preimage (one or two choices) works and we have a stochastic solution $Q$. (Associate $p',q'>0$ with sum $1$ and difference $t$.) If not, there is no such $Q$. In the given generality we have a negative answer.
This inexistence was a consequence of the fact that $g_2:(-1,1)\to(-1,2)$  is not surjective. For a general $C>1$, the function $g_C:(-1,1)\to(-1,1)$ is also non-surjective. (Its restriction $[0,1)\to[0,1)$ is increasing and bijective, but on $(-1,0)$ the image keeps a positive distance to $-1$, since there is at least one positive term in the defining sum.) The same argument applies.

Note: Similar "symmetric" examples can be written considering a transition matrix on states building a group $(A,+)$, with fixed constants $p_a$ corresponding to the probable transition from a state $x\in A$ to the state $x+a$. The above example is obtained for $A=\Bbb Z/2$.
