Given a vector space $V$ and endomorphism $M : V \to V$ we can define the exponential of $M$: $\exp(M) = \sum_{k=0}^{\infty} \frac{1}{k!}M^k$. This has the property that it commutes with conjugation in the sense that $g \exp(M) g^{-1} = \exp(g M g^{-1})$ for $g : V \to V$ invertible.
I am wondering if the following generalization is true: Let $V$ and $W$ be vector spaces, $M : V \to V$ and $N : W \to W$ be endomorphisms, and $T : V \to W$ a linear transformation (not necessarily invertible) such that $T M = NT$. I would like to know if it then follows that $T \exp(M) = \exp(N) T$.
Expanding my comment $$T\exp(M)=T\left(\sum_{k\ge 0}\frac{M^k}{k!}\right)=\sum_{k\ge 0}\frac{TM^k}{k!}=\sum_{k\ge 0}\frac{N^kT}{k!}=\left(\sum_{k\ge 0}\frac{N^k}{k!}\right)T=\exp(N)T$$