**Appendix A. Formulas for** *<sup>P</sup><sup>m</sup>* **and** *<sup>F</sup>* **for the Case** *<sup>p</sup>***<sup>1</sup> <sup>=</sup>** *<sup>p</sup>***<sup>2</sup> <sup>=</sup>** *<sup>p</sup>***<sup>3</sup> <sup>=</sup>** *<sup>p</sup>*

Formulas for the fidelity and the success probability as the number of channels in indefinite causal order increases when *p*<sup>1</sup> = *p*<sup>2</sup> = *p*<sup>3</sup> = *p* have been obtained. For the case *N* = 2, when |*ψm* = |*ϕ*<sup>−</sup> *m* the outcomes are:

$$\mathcal{F}\_2 = \frac{1}{3'} \qquad \mathcal{P}\_m = 6p^2 \tag{A1}$$

and for the case when <sup>|</sup>*ψm* <sup>=</sup> <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>*, the expressions become:

$$\mathcal{F}\_2 = \frac{6p^2 - 4p + 1}{1 - 6p^2}, \qquad \mathcal{P}\_m = 1 - 6p^2. \tag{A2}$$

For the case *N* = 3, when |*ψm* = |*ϕ*<sup>−</sup> *<sup>m</sup>* the outcomes are:

$$\mathcal{F}\_3 = \frac{1}{3} + 2p, \qquad \mathcal{P}\_m = 2p^2 \tag{A3}$$

and for the case when <sup>|</sup>*ψm* <sup>=</sup> <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>*, we get:

$$\mathcal{F}\_3 = \frac{-76p^3 + 54p^2 - 18p + 3}{96p^3 - 54p^2 + 3}, \qquad \mathcal{P}\_m = 1 - 18p^2 + 32p^3. \tag{A4}$$

For the case *N* = 4, when |*ψm* = |*ϕ*<sup>−</sup> *<sup>m</sup>* we get P*<sup>m</sup>* = 0, thus F<sup>4</sup> becomes undefined in such case, while for the case when <sup>|</sup>*ψm* <sup>=</sup> <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>*, we get the expressions:

$$\mathcal{F}\_4 = \frac{360p^4 - 304p^3 + 108p^2 - 24p + 3}{-408p^4 + 384p^3 - 108p^2 + 3}, \qquad \mathcal{P}\_{\mathfrak{M}} = 1 - 36p^2 + 128p^3 - 136p^4. \tag{A5}$$

#### **Appendix B. Formulas for** *<sup>P</sup>***ff,***{pi} <sup>m</sup>***,***<sup>N</sup>* **and** *<sup>F</sup>* **for the Case** *<sup>p</sup>***<sup>0</sup> <sup>=</sup> <sup>0</sup>**

In this section, formulas for <sup>F</sup> and <sup>P</sup>ff,{*pi*} *<sup>m</sup>*,*<sup>N</sup>* when the entangled state has different values for *p*1, *p*<sup>2</sup> and *p*<sup>3</sup> (note they are restricted to the frontal face *p*<sup>0</sup> = 0 of the parametric space) and the measurement of the control state is either <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>* or |*ϕ*<sup>−</sup> *<sup>m</sup>*. In those results, the angles *θ* and *φ* corresponds to the state being teleported (|*ψ* <sup>=</sup> cos *<sup>θ</sup>* <sup>2</sup> <sup>|</sup>0 <sup>+</sup> sin *<sup>θ</sup>* <sup>2</sup> *<sup>e</sup>i<sup>φ</sup>* <sup>|</sup>1), thus meaning a dependence of those values on this state. For the case *<sup>N</sup>* <sup>=</sup> 2, with the privileged measurement state as <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>*, the expressions become:

$$
\begin{array}{rcl}
\mathcal{F}\_2 & = & 1 \\
\end{array}
\tag{A6}
$$

$$\mathcal{P}\_{m,N=2}^{\text{tt},\{p\_i\}} = \ \ \ p\_1^2 + p\_2^2 + p\_3^2 \tag{A7}$$

and with the privileged state as |*ϕ*<sup>−</sup> *<sup>m</sup>*, the corresponding expressions are:

$$\mathcal{F}\_2 = \frac{1}{2\mathcal{P}\_{m,N=2}^{\text{ff},\{p\_i\}}} \Big( 2p\_1 p\_2 (1 + \cos 2\theta) + p\_3 (p\_1 + p\_2)(1 - \cos 2\theta) \Big) \tag{A8}$$

$$+2p\_3(p\_2 - p\_1)\sin^2\theta\cos2\phi\Big)$$

$$\mathcal{P}\_{m,N=2}^{\text{ft},\{p\_i\}} = 2(p\_1p\_2 + p\_2p\_3 + p\_3p\_1). \tag{A9}$$

For the case *<sup>N</sup>* <sup>=</sup> 3, with the privileged measurement state as <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>*, the outcomes are:

$$\mathcal{F}\_3 = \frac{1}{12\mathcal{P}\_{m,N=3}^{\text{fit, }\{p\_i\}}} \left( (3(p\_1^3 + p\_2^3 + 2p\_3^3) + p\_1(p\_2^2 + p\_3^2) + p\_2(p\_1^2 + p\_3^2))(1 - \cos 2\theta) \right) \tag{A10}$$

$$1 + 2p\_3(p\_1^2 + p\_2^2)(1 + \cos 2\theta) + 2(3(p\_1^3 - p\_2^3) + p\_1(p\_2^2 + p\_3^2) - p\_2(p\_1^2 + p\_3^2))\sin^2\theta\cos 2\phi$$

$$\mathcal{P}\_{m,N=3}^{\text{eff.}\{p\_i\}} = \quad p\_1^3 + p\_2^3 + p\_3^3 + \frac{1}{3}(p\_1^2(p\_2 + p\_3) + p\_2^2(p\_1 + p\_3) + p\_3^2(p\_1 + p\_2))\tag{A11}$$

while, with the privileged state as |*ϕ*<sup>−</sup> *<sup>m</sup>*, they become:

$$\mathcal{F}\_{\frac{3}{2}} = \mathbf{1} \tag{A12}$$

$$\mathcal{P}\_{m,N=3}^{\text{ff.}\{p\_i^{\prime}\}} = \; \not\models p\_1 p\_2 p\_3. \tag{A13}$$

Finally, for the case *<sup>N</sup>* <sup>=</sup> 4, with the privileged measurement state as <sup>|</sup>*ϕ*<sup>+</sup> *<sup>m</sup>*, then:

$$\mathcal{F}\_4 \quad = \quad 1 \tag{A14}$$

$$\mathcal{P}\_{m,N=4}^{\text{ff.}\{p\_i\}} = \quad p\_1^4 + p\_2^4 + p\_3^4 + \frac{2}{3}(p\_1^2 p\_2^2 + p\_1^2 p\_3^2 + p\_2^2 p\_3^2) \tag{A15}$$

and if the privileged state is |*ϕ*<sup>−</sup> *<sup>m</sup>*, then we get <sup>P</sup>ff,{*pi*} *<sup>m</sup>*,*N*=<sup>4</sup> = 0, thus F gets undetermined.
