$M_n=M_n(\mathbb C)$ is the $n\times n$ complex matrix algebra and $M_n(C(X))$ is the algebra of continuous functions from $X$ to $M_n(\mathbb C)$.
There is a trivial homomorphism $M_n\to M_n(C(X))$ which maps a matrix to a constant function taking value in $M_n$. Is every nonzero homomorphism unitarily equivalent to the trivial one?
Denote by $e_i$ the $n$ orthogonal equivalent projections in $M_n$. if $\varphi:M_n\to M_n(C(X))$ is a nonzero homomorphism, $\varphi(e_i)=p_i$ are equivalent projections such that $p_1+\cdots +p_n=1_n\otimes 1_X$. Assume $p_i$ are not stably equivalent to the trivial projections $e_i\otimes 1_X$ (though stably equivalent projections are not necessarily equivalent), then $[p_i]\neq [e_i\otimes 1_X]$. Since $n[p_i]=n[e_i\otimes 1_X]$, it follows that $K_0(C(X))$ can not be torsion-free.
Take this as an example. Let $X$ be the CW complex obtained by attaching the 2-dimensional disc $D^2=\{z\in \mathbb C:|z|\leq 1\}$ to the circle $S^1=\{z\in \mathbb C:|z|=1\}$ via a degree 2 map. It is known that $K_0(C(X))\simeq \mathbb Z_2\oplus \mathbb Z$, which can be computed using
$$\require{AMScd} \begin{CD} K_0(S^2\mathbb C)\simeq \mathbb Z @>>> K_0(C(X)) @>>> \mathbb Z\simeq K_0(C(S^1))\\ @AAA @. @VVV \\ K_1(C(S^1))\simeq \mathbb Z @<<< K_1(C(X)) @<<< 0=K_1(S^2\mathbb C) \end{CD}$$
and
$$\begin{CD} K_0(S^2\mathbb C) @>>> K_0(C(S^2))\\ @VVV @VVV \\ K_0(C(X)) @>>> K_0(C(X)\oplus \mathbb C). \end{CD}$$
One computes that the generator in $K_0(S^2\mathbb C)$ is represented by $\left[\left(\begin{array}{}1&0\\0&0\end{array}\right)\right]-\left[\left(\begin{array}{}1-zz^*&-z(1-z^*z)^{1/2}\\-z^*(1-z^*z)^{1/2}&z^*z\end{array}\right)\right]\in K_0(C(S^2))$.
Hence $\left[\left(\begin{array}{}1&0\\0&0\end{array}\right)\right]=(0,1)$ and $\left[\left(\begin{array}{}1-zz^*&-z(1-z^*z)^{1/2}\\-z^*(1-z^*z)^{1/2}&z^*z\end{array}\right)\right]=(1,1)$ in $K_0(C(X))\simeq \mathbb Z_2\oplus \mathbb Z$. It then follows $\left[\left(\begin{array}{}1-zz^*&-z(1-z^*z)^{1/2}\\-z^*(1-z^*z)^{1/2}&z^*z\end{array}\right)\right]=\left[\left(\begin{array}{}zz^*&z(1-z^*z)^{1/2}\\z^*(1-z^*z)^{1/2}&1-z^*z\end{array}\right)\right]$.
I don't think they are unitarily equivalent but I have no proof.